[{"Name":"Stoichiometry","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"9m 21s","ChapterTopicVideoID":22945,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/22945.jpeg","UploadDate":"2020-09-23T14:19:34.3900000","DurationForVideoObject":"PT9M21S","Description":null,"MetaTitle":"Exercise 1 - Stoichiometry: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Stoichiometry practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/general-chemistry/chemical-reactions/stoichiometry/vid23791","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"Hi. We\u0027re going to balance the following equations."},{"Start":"00:03.510 ","End":"00:08.295","Text":"In a, we have magnesium plus oxygen,"},{"Start":"00:08.295 ","End":"00:10.695","Text":"gives us magnesium oxide."},{"Start":"00:10.695 ","End":"00:14.520","Text":"We\u0027re going to divide the equation to the reactants and the products side."},{"Start":"00:14.520 ","End":"00:19.530","Text":"In the reactant side, we have 1 magnesium and 2 oxygens."},{"Start":"00:19.530 ","End":"00:24.930","Text":"In the product side, we have 1 magnesium and 1 oxygen."},{"Start":"00:24.930 ","End":"00:26.760","Text":"Of course, we\u0027re not going to touch them magnesium"},{"Start":"00:26.760 ","End":"00:28.485","Text":"at this point because they are balanced,"},{"Start":"00:28.485 ","End":"00:30.330","Text":"and the oxygens are not."},{"Start":"00:30.330 ","End":"00:32.640","Text":"Therefore, instead of 1 oxygen here,"},{"Start":"00:32.640 ","End":"00:34.595","Text":"we\u0027re going to multiply this by 2."},{"Start":"00:34.595 ","End":"00:38.750","Text":"Therefore, we will add the coefficient 2 to the magnesium oxide."},{"Start":"00:38.750 ","End":"00:40.970","Text":"Now that we have 2 magnesium oxide,"},{"Start":"00:40.970 ","End":"00:42.335","Text":"we have 2 oxygens,"},{"Start":"00:42.335 ","End":"00:44.675","Text":"and we also have 2 magnesiums."},{"Start":"00:44.675 ","End":"00:48.065","Text":"We need to change our magnesium to 2. Now we will compare."},{"Start":"00:48.065 ","End":"00:51.215","Text":"In the reactant side, we have 1 magnesium,"},{"Start":"00:51.215 ","End":"00:53.330","Text":"and 2 oxygens, and then in the product side,"},{"Start":"00:53.330 ","End":"00:55.460","Text":"we have 2 magnesiums and 2 oxygens."},{"Start":"00:55.460 ","End":"00:58.760","Text":"Our next step is to change our magnesium to 2."},{"Start":"00:58.760 ","End":"01:01.280","Text":"For this purpose, we will multiply the magnesium by 2."},{"Start":"01:01.280 ","End":"01:04.475","Text":"Now we have 2 magnesiums and 2 oxygens on the reactant side,"},{"Start":"01:04.475 ","End":"01:07.210","Text":"and 2 magnesiums and 2 oxygens on the product side."},{"Start":"01:07.210 ","End":"01:08.990","Text":"Now the equation is balanced."},{"Start":"01:08.990 ","End":"01:14.540","Text":"Now, we\u0027ll go on to b. Ammonia plus oxygen gives us nitrogen dioxide plus water."},{"Start":"01:14.540 ","End":"01:16.840","Text":"We\u0027re going to write this down here."},{"Start":"01:16.840 ","End":"01:26.245","Text":"Ammonia plus oxygen gives us nitrogen dioxide plus water."},{"Start":"01:26.245 ","End":"01:29.870","Text":"Again, we\u0027re going to divide this into the reactant side and the product side."},{"Start":"01:29.870 ","End":"01:32.255","Text":"On the reactant side, we have 1 nitrogen,"},{"Start":"01:32.255 ","End":"01:37.385","Text":"3 hydrogens from here, and 2 oxygens."},{"Start":"01:37.385 ","End":"01:40.790","Text":"On the product side, we also have 1 nitrogen,"},{"Start":"01:40.790 ","End":"01:44.960","Text":"2 hydrogens, and we have 3 oxygens,"},{"Start":"01:44.960 ","End":"01:49.180","Text":"1 from the water, and 2 for the nitrogen dioxide, so 3 oxygens."},{"Start":"01:49.180 ","End":"01:52.490","Text":"Right now, our nitrogens are balanced,"},{"Start":"01:52.490 ","End":"01:53.810","Text":"therefore, we\u0027re not going to touch them."},{"Start":"01:53.810 ","End":"01:57.080","Text":"We have to decide if to change the hydrogens or the oxygens."},{"Start":"01:57.080 ","End":"02:01.129","Text":"So since we have free oxygen in our reactants,"},{"Start":"02:01.129 ","End":"02:03.350","Text":"we\u0027re not going to touch the oxygens right now,"},{"Start":"02:03.350 ","End":"02:06.455","Text":"we\u0027re going to leave them till last because it\u0027s going to be very easy to balance."},{"Start":"02:06.455 ","End":"02:08.645","Text":"Now we\u0027ll go on to the hydrogens."},{"Start":"02:08.645 ","End":"02:12.310","Text":"We have 2 in our products and 3 in our reactants."},{"Start":"02:12.310 ","End":"02:16.960","Text":"What we\u0027re going to do, is we\u0027re going to multiply the hydrogens by 2 in our reactants,"},{"Start":"02:16.960 ","End":"02:19.709","Text":"that way we will get 6 hydrogens,"},{"Start":"02:19.709 ","End":"02:22.940","Text":"and in our products, we will multiply them by 3."},{"Start":"02:22.940 ","End":"02:27.930","Text":"This is plus 3 H_2O."},{"Start":"02:27.930 ","End":"02:30.120","Text":"That way, we\u0027ll also have 6."},{"Start":"02:30.120 ","End":"02:33.940","Text":"First of all, let\u0027s change our hydrogens."},{"Start":"02:34.490 ","End":"02:37.800","Text":"As we said, we\u0027re going to have 6 hydrogens here,"},{"Start":"02:37.800 ","End":"02:39.240","Text":"and 6 hydrogens here."},{"Start":"02:39.240 ","End":"02:41.960","Text":"But that\u0027s not it. When we change the hydrogens,"},{"Start":"02:41.960 ","End":"02:45.370","Text":"we also change the nitrogen on this side."},{"Start":"02:45.370 ","End":"02:49.850","Text":"We changed it to 2 since we have 2 NH_3 and on this side,"},{"Start":"02:49.850 ","End":"02:50.975","Text":"we changed the hydrogens,"},{"Start":"02:50.975 ","End":"02:52.505","Text":"but we also changed the oxygen."},{"Start":"02:52.505 ","End":"02:55.190","Text":"Now we have 2 oxygens from the nitrogen dioxide,"},{"Start":"02:55.190 ","End":"02:57.530","Text":"and 3 oxygens from the water."},{"Start":"02:57.530 ","End":"02:59.060","Text":"Instead of 3 oxygens,"},{"Start":"02:59.060 ","End":"03:02.150","Text":"we have 5 oxygens at this point."},{"Start":"03:02.150 ","End":"03:05.075","Text":"Now if we compare, we have 6 hydrogens in the reactants,"},{"Start":"03:05.075 ","End":"03:06.755","Text":"6 hydrogens in the product."},{"Start":"03:06.755 ","End":"03:08.975","Text":"But we have 2 nitrogens in the reactants,"},{"Start":"03:08.975 ","End":"03:10.475","Text":"and 1 in the product."},{"Start":"03:10.475 ","End":"03:15.575","Text":"Therefore, we\u0027re going to multiply our nitrogen dioxide by 2,"},{"Start":"03:15.575 ","End":"03:17.450","Text":"so we get 2 nitrogens."},{"Start":"03:17.450 ","End":"03:20.150","Text":"We\u0027re going to change the nitrogens to 2."},{"Start":"03:20.150 ","End":"03:24.980","Text":"Also, we change the number of oxygens because we have 4 from the nitrogen dioxide,"},{"Start":"03:24.980 ","End":"03:29.040","Text":"2 times 2, plus another 3 from our water."},{"Start":"03:29.040 ","End":"03:32.025","Text":"We have a total of 7 oxygens."},{"Start":"03:32.025 ","End":"03:36.180","Text":"We\u0027re going to change this to 7. There we go."},{"Start":"03:36.180 ","End":"03:39.840","Text":"Now we have 2 nitrogens in the reactants and 2 nitrogens in our product,"},{"Start":"03:39.840 ","End":"03:43.470","Text":"6 hydrogens in the reactants and 6 hydrogens in the products,"},{"Start":"03:43.470 ","End":"03:48.600","Text":"and we have 2 oxygens in our reactants and 7 oxygens in the products."},{"Start":"03:48.600 ","End":"03:51.620","Text":"What we\u0027re going to do in order to balance them out,"},{"Start":"03:51.620 ","End":"03:54.320","Text":"2 times 7/2 equals 7."},{"Start":"03:54.320 ","End":"03:57.755","Text":"Therefore, we will multiply our oxygen by 7/2."},{"Start":"03:57.755 ","End":"03:59.915","Text":"That way, instead of 2,"},{"Start":"03:59.915 ","End":"04:02.480","Text":"we now have 7 oxygens."},{"Start":"04:02.480 ","End":"04:06.000","Text":"Now, our equation is balanced."},{"Start":"04:06.000 ","End":"04:10.890","Text":"The balanced equation for b is 2 ammonia plus 7/2 oxygen,"},{"Start":"04:10.890 ","End":"04:14.360","Text":"gives us 2 nitrogen dioxide plus 3 water."},{"Start":"04:14.360 ","End":"04:16.940","Text":"Now, you can leave the coefficient as a fraction."},{"Start":"04:16.940 ","End":"04:18.455","Text":"However, if it bothers you,"},{"Start":"04:18.455 ","End":"04:20.390","Text":"you can multiply everything by 2,"},{"Start":"04:20.390 ","End":"04:24.140","Text":"and now you will get 4 ammonia plus 7 oxygen,"},{"Start":"04:24.140 ","End":"04:27.880","Text":"gives you 4 nitrogen dioxide plus 6 water."},{"Start":"04:27.880 ","End":"04:29.205","Text":"Whatever way you like it,"},{"Start":"04:29.205 ","End":"04:31.820","Text":"both ways are perfect."},{"Start":"04:31.820 ","End":"04:35.035","Text":"Now we\u0027re going to balance c. Ethane,"},{"Start":"04:35.035 ","End":"04:38.140","Text":"C_2H_6 plus oxygen,"},{"Start":"04:38.140 ","End":"04:41.124","Text":"O_2, gives us carbon dioxide,"},{"Start":"04:41.124 ","End":"04:44.575","Text":"CO_2, plus water, H_2O."},{"Start":"04:44.575 ","End":"04:47.915","Text":"We\u0027re going to divide this again into the reactant side and product side."},{"Start":"04:47.915 ","End":"04:51.050","Text":"In the reactant side, we can see we have 2 carbons,"},{"Start":"04:51.050 ","End":"04:53.790","Text":"we have 6 hydrogens,"},{"Start":"04:53.790 ","End":"04:56.360","Text":"and we have 2 oxygens."},{"Start":"04:56.360 ","End":"04:58.940","Text":"On the product side, we have 1 carbon,"},{"Start":"04:58.940 ","End":"05:06.020","Text":"2 hydrogens, and we have 2 oxygens from the carbon dioxide and 1 from the water,"},{"Start":"05:06.020 ","End":"05:08.300","Text":"so we have 3 oxygens."},{"Start":"05:08.300 ","End":"05:10.670","Text":"Then, we\u0027re going to start with a carbon."},{"Start":"05:10.670 ","End":"05:12.410","Text":"We have 2 carbons on the reactant side,"},{"Start":"05:12.410 ","End":"05:14.270","Text":"1 carbon on the product side."},{"Start":"05:14.270 ","End":"05:17.030","Text":"We\u0027re going to multiply the carbon dioxide by 2."},{"Start":"05:17.030 ","End":"05:20.360","Text":"This way, we\u0027re going to get 2 carbons on the product side,"},{"Start":"05:20.360 ","End":"05:24.170","Text":"and also our oxygen will change since now we have 4 oxygens,"},{"Start":"05:24.170 ","End":"05:27.255","Text":"2 times 2 from the carbon dioxide,"},{"Start":"05:27.255 ","End":"05:29.640","Text":"plus 1 oxygen from the water,"},{"Start":"05:29.640 ","End":"05:32.415","Text":"so we have 5 oxygens instead of 3."},{"Start":"05:32.415 ","End":"05:35.670","Text":"We\u0027re going to erase the 3 and write down 5."},{"Start":"05:35.670 ","End":"05:38.420","Text":"In the next step, we\u0027re going to balance"},{"Start":"05:38.420 ","End":"05:43.235","Text":"the hydrogens since we have a free oxygen and it\u0027s very easy to balance at the end."},{"Start":"05:43.235 ","End":"05:45.980","Text":"We have 6 hydrogens on the reactant side and on the product side,"},{"Start":"05:45.980 ","End":"05:47.375","Text":"we have 2 hydrogens."},{"Start":"05:47.375 ","End":"05:50.725","Text":"What we\u0027re going to do, is we\u0027re going to multiply our water by 3."},{"Start":"05:50.725 ","End":"05:53.820","Text":"That way we get 6 hydrogens, 2 times 3."},{"Start":"05:53.820 ","End":"05:56.220","Text":"Our hydrogen changes 2 to 6."},{"Start":"05:56.220 ","End":"05:58.565","Text":"As you can see, we also changed our oxygens."},{"Start":"05:58.565 ","End":"06:01.320","Text":"We have 3 oxygens from the water,"},{"Start":"06:01.320 ","End":"06:04.020","Text":"plus 4 oxygens from the carbon dioxide,"},{"Start":"06:04.020 ","End":"06:06.720","Text":"so we have 7 oxygens in total."},{"Start":"06:06.720 ","End":"06:08.340","Text":"We\u0027re just going to write 7."},{"Start":"06:08.340 ","End":"06:10.155","Text":"Now our carbons are balanced,"},{"Start":"06:10.155 ","End":"06:11.420","Text":"our hydrogens are balanced,"},{"Start":"06:11.420 ","End":"06:13.070","Text":"and we only have to balance our oxygen."},{"Start":"06:13.070 ","End":"06:15.830","Text":"In the product, we have 7 oxygens,"},{"Start":"06:15.830 ","End":"06:18.870","Text":"and then in the reactants, we have 2 oxygens."},{"Start":"06:18.870 ","End":"06:21.650","Text":"What we\u0027re going to do in order to change the 2 into 7,"},{"Start":"06:21.650 ","End":"06:28.425","Text":"we\u0027re just going to multiply our oxygen by 7/2 because 7/2 times 2 equals 7."},{"Start":"06:28.425 ","End":"06:30.480","Text":"Let\u0027s change this 2 to 7."},{"Start":"06:30.480 ","End":"06:32.030","Text":"Now we see that our carbons,"},{"Start":"06:32.030 ","End":"06:34.536","Text":"hydrogens, and oxygens are balanced."},{"Start":"06:34.536 ","End":"06:36.470","Text":"This is our balanced equation."},{"Start":"06:36.470 ","End":"06:40.205","Text":"Now, you can leave the coefficient as a fraction if you like."},{"Start":"06:40.205 ","End":"06:44.405","Text":"If you don\u0027t like, you can also multiply the whole equation by 2."},{"Start":"06:44.405 ","End":"06:46.100","Text":"If you multiply the whole equation by 2,"},{"Start":"06:46.100 ","End":"06:52.805","Text":"we will get 2 ethane plus 7 oxygen gives us 4 carbon dioxide plus 6 water."},{"Start":"06:52.805 ","End":"06:54.590","Text":"Both ways are correct."},{"Start":"06:54.590 ","End":"07:01.730","Text":"That is our final answer for c. Now we\u0027re balancing d. Lead oxide plus"},{"Start":"07:01.730 ","End":"07:11.025","Text":"ammonia gives us lead plus nitrogen plus water."},{"Start":"07:11.025 ","End":"07:14.600","Text":"We\u0027re going to divide this into the reactant side and the product side."},{"Start":"07:14.600 ","End":"07:17.945","Text":"In our reactant side, we have 1 times lead,"},{"Start":"07:17.945 ","End":"07:24.795","Text":"1 times nitrogen, 3 hydrogens, and 1 oxygen."},{"Start":"07:24.795 ","End":"07:28.155","Text":"In our product side, we have 1 lead,"},{"Start":"07:28.155 ","End":"07:36.400","Text":"2 nitrogens, 2 hydrogens, and 1 oxygen."},{"Start":"07:36.470 ","End":"07:38.870","Text":"Right now the lead is balanced,"},{"Start":"07:38.870 ","End":"07:40.225","Text":"so I\u0027m not going to touch it."},{"Start":"07:40.225 ","End":"07:42.425","Text":"I\u0027m going to begin with the nitrogens."},{"Start":"07:42.425 ","End":"07:47.835","Text":"Here we can see that we have 2 nitrogens in the product and 1 nitrogen in our reactant,"},{"Start":"07:47.835 ","End":"07:51.410","Text":"therefore, we\u0027re going to multiply the ammonia by 2."},{"Start":"07:51.410 ","End":"07:55.060","Text":"This way, we\u0027re going to change our nitrogen to 2."},{"Start":"07:55.060 ","End":"07:58.595","Text":"We can also see that this changes the hydrogens to 6,"},{"Start":"07:58.595 ","End":"08:01.040","Text":"since the hydrogens are 3 times 2,"},{"Start":"08:01.040 ","End":"08:02.405","Text":"so they equal 6."},{"Start":"08:02.405 ","End":"08:06.520","Text":"We\u0027re going to change that 2, hydrogens are 6."},{"Start":"08:06.520 ","End":"08:08.820","Text":"Now our lead is 1 and 1,"},{"Start":"08:08.820 ","End":"08:10.455","Text":"our nitrogen is 2 and 2,"},{"Start":"08:10.455 ","End":"08:14.675","Text":"and our hydrogens are 6 and 2 on the product side."},{"Start":"08:14.675 ","End":"08:16.085","Text":"In order to change this,"},{"Start":"08:16.085 ","End":"08:18.490","Text":"we\u0027re going to multiply the water by 3."},{"Start":"08:18.490 ","End":"08:21.450","Text":"That gives us 6 hydrogens because it\u0027s 2 times 3,"},{"Start":"08:21.450 ","End":"08:24.470","Text":"so we\u0027re going to change the hydrogens to 6."},{"Start":"08:24.470 ","End":"08:27.455","Text":"But it also changes our oxygens to 3."},{"Start":"08:27.455 ","End":"08:29.935","Text":"We have 3 oxygens instead of 1."},{"Start":"08:29.935 ","End":"08:32.860","Text":"We\u0027re going to change 1 oxygen to 3."},{"Start":"08:32.860 ","End":"08:35.915","Text":"Now, everything is balanced except for oxygen."},{"Start":"08:35.915 ","End":"08:38.885","Text":"We have 3 in the products and 1 in our reactants."},{"Start":"08:38.885 ","End":"08:42.110","Text":"What we need to do is multiply lead oxide by 3,"},{"Start":"08:42.110 ","End":"08:44.435","Text":"that way we get 3 oxygens,"},{"Start":"08:44.435 ","End":"08:46.905","Text":"we can change the oxygen to 3."},{"Start":"08:46.905 ","End":"08:49.520","Text":"Now, that we multiplied the lead oxide by 3,"},{"Start":"08:49.520 ","End":"08:51.485","Text":"we need to change our lead also to 3."},{"Start":"08:51.485 ","End":"08:54.440","Text":"I\u0027m going to change on 1 lead to 3."},{"Start":"08:54.440 ","End":"08:57.125","Text":"Now when we take a look at the reactants and the products,"},{"Start":"08:57.125 ","End":"08:59.225","Text":"we see that there\u0027s a change in the lead."},{"Start":"08:59.225 ","End":"09:04.095","Text":"There are 3 leads in the reactants and 1 in the product."},{"Start":"09:04.095 ","End":"09:07.685","Text":"We\u0027re just going to take the lead and multiply it by 3."},{"Start":"09:07.685 ","End":"09:09.035","Text":"That way, 3 lead."},{"Start":"09:09.035 ","End":"09:11.825","Text":"Now you can see everything is balanced, the lead,"},{"Start":"09:11.825 ","End":"09:13.310","Text":"the nitrogen, the hydrogens,"},{"Start":"09:13.310 ","End":"09:15.025","Text":"and the oxygens are balanced."},{"Start":"09:15.025 ","End":"09:19.700","Text":"This is our balanced equation for d. That is our final answer."},{"Start":"09:19.700 ","End":"09:22.380","Text":"Thank you very much for watching."}],"ID":23791},{"Watched":false,"Name":"Exercise 2","Duration":"7m 25s","ChapterTopicVideoID":22946,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.170","Text":"Hi, we\u0027re going to balance the following equations."},{"Start":"00:03.170 ","End":"00:08.100","Text":"In a, we have phosphoric acid plus calcium oxide,"},{"Start":"00:08.100 ","End":"00:14.535","Text":"gives us tricalcium phosphate plus water."},{"Start":"00:14.535 ","End":"00:18.210","Text":"We\u0027re going to divide this in the reactants and the products."},{"Start":"00:18.210 ","End":"00:21.970","Text":"Then the reactants, we can see we have 1 calcium,"},{"Start":"00:23.090 ","End":"00:28.065","Text":"1 oxygen, 3 hydrogens,"},{"Start":"00:28.065 ","End":"00:31.180","Text":"and 1 phosphate ion."},{"Start":"00:32.680 ","End":"00:35.510","Text":"Now, when we have polyatomic ions,"},{"Start":"00:35.510 ","End":"00:37.070","Text":"we leave them as a unit,"},{"Start":"00:37.070 ","End":"00:40.310","Text":"so we don\u0027t separate the phosphorus from the oxygen."},{"Start":"00:40.310 ","End":"00:42.995","Text":"We leave them together. It\u0027s 1 phosphate ion."},{"Start":"00:42.995 ","End":"00:44.780","Text":"Now if we look here at the product,"},{"Start":"00:44.780 ","End":"00:46.235","Text":"we have 3 calcium,"},{"Start":"00:46.235 ","End":"00:51.250","Text":"1 oxygen, we have 2 hydrogens right over here."},{"Start":"00:51.250 ","End":"00:54.430","Text":"We have 2 phosphate ions."},{"Start":"00:56.000 ","End":"00:58.070","Text":"Let\u0027s start with our calcium."},{"Start":"00:58.070 ","End":"01:02.105","Text":"We see that in our reactants we have 1 calcium and then our products 3."},{"Start":"01:02.105 ","End":"01:05.300","Text":"We have to multiply our calcium by 3."},{"Start":"01:05.300 ","End":"01:08.900","Text":"We\u0027re going to multiply the calcium oxide by 3."},{"Start":"01:08.900 ","End":"01:12.560","Text":"This is going to change our calcium to"},{"Start":"01:12.560 ","End":"01:18.890","Text":"3 and this is also going to change our oxygen to 3 because oxygen is also times 3."},{"Start":"01:19.250 ","End":"01:22.015","Text":"Now our calciums are balanced."},{"Start":"01:22.015 ","End":"01:23.530","Text":"It\u0027ll go onto the oxygen,"},{"Start":"01:23.530 ","End":"01:26.800","Text":"here we have 3 and here we have 1 oxygen and the hydrogens,"},{"Start":"01:26.800 ","End":"01:28.945","Text":"we have 3 hydrogens in the reactant."},{"Start":"01:28.945 ","End":"01:30.745","Text":"We have 3 hydrogens in the reactants,"},{"Start":"01:30.745 ","End":"01:32.690","Text":"and 2 hydrogens in our products."},{"Start":"01:32.690 ","End":"01:36.465","Text":"The phosphate ion we have 1 here and 2 here."},{"Start":"01:36.465 ","End":"01:39.685","Text":"Let\u0027s leave our hydrogens and oxygens to last."},{"Start":"01:39.685 ","End":"01:41.680","Text":"Let\u0027s go for the phosphate ion."},{"Start":"01:41.680 ","End":"01:43.630","Text":"In our reactants, we have 1."},{"Start":"01:43.630 ","End":"01:47.590","Text":"In our products, we have 2, so we have to multiply the phosphate ion by 2."},{"Start":"01:47.590 ","End":"01:51.310","Text":"I\u0027m going to take us phosphoric acid and multiply it by 2."},{"Start":"01:51.310 ","End":"01:53.520","Text":"That way, the phosphate ion,"},{"Start":"01:53.520 ","End":"01:55.945","Text":"I\u0027m just going to change to 2."},{"Start":"01:55.945 ","End":"01:58.520","Text":"But we also change the amount of hydrogens."},{"Start":"01:58.520 ","End":"02:00.770","Text":"Now we have 3 times 2 hydrogens,"},{"Start":"02:00.770 ","End":"02:02.690","Text":"meaning we have 6 hydrogens."},{"Start":"02:02.690 ","End":"02:05.930","Text":"We\u0027re going to change our hydrogens to 6."},{"Start":"02:05.930 ","End":"02:08.960","Text":"Now if we look, we can see that our calcium is"},{"Start":"02:08.960 ","End":"02:12.185","Text":"balanced and our phosphate is also balanced."},{"Start":"02:12.185 ","End":"02:15.124","Text":"We\u0027re going to go on to the oxygens and the hydrogens."},{"Start":"02:15.124 ","End":"02:18.605","Text":"In our reactants, we have 3 oxygens and 6 hydrogens,"},{"Start":"02:18.605 ","End":"02:21.875","Text":"and in our products, we have 1 oxygen and 2 hydrogens."},{"Start":"02:21.875 ","End":"02:25.280","Text":"We can fix this if we multiply our water by 3."},{"Start":"02:25.280 ","End":"02:28.160","Text":"That way the oxygen is going to change to 3,"},{"Start":"02:28.160 ","End":"02:31.850","Text":"and the hydrogens will change to 6 because it\u0027s 2 times 3 equals 6,"},{"Start":"02:31.850 ","End":"02:35.820","Text":"so 3 oxygens and 6 hydrogens."},{"Start":"02:35.820 ","End":"02:38.370","Text":"Now our equation is balanced."},{"Start":"02:38.370 ","End":"02:40.685","Text":"That\u0027s the balanced equation for a."},{"Start":"02:40.685 ","End":"02:45.575","Text":"Now we will balance b and b we have nitric oxide plus oxygen,"},{"Start":"02:45.575 ","End":"02:48.005","Text":"gives us nitrogen dioxide."},{"Start":"02:48.005 ","End":"02:51.170","Text":"We\u0027re going to divide this into the reactant side and the product side."},{"Start":"02:51.170 ","End":"02:54.980","Text":"In the reactant side, we can see that we have 1 nitrogen and we have"},{"Start":"02:54.980 ","End":"03:00.935","Text":"3 oxygens but the oxygens are 1 from the nitric oxide and 2 from oxygen."},{"Start":"03:00.935 ","End":"03:05.025","Text":"In our product side, we have 1 nitrogen and 2 oxygens."},{"Start":"03:05.025 ","End":"03:09.590","Text":"Right now the nitrogens are balanced 1 and 1 and the oxygens differ."},{"Start":"03:09.590 ","End":"03:11.090","Text":"If we look at our equation,"},{"Start":"03:11.090 ","End":"03:16.190","Text":"you can see that the easiest way to balance the oxygens is to"},{"Start":"03:16.190 ","End":"03:21.890","Text":"multiply our nitric oxide by 2 and our nitrogen dioxide by 2."},{"Start":"03:21.890 ","End":"03:23.975","Text":"This way, let\u0027s look at what we did."},{"Start":"03:23.975 ","End":"03:27.740","Text":"This way we change our nitrogens in the reactant side to 2."},{"Start":"03:27.740 ","End":"03:30.440","Text":"We\u0027re going to change the nitrogens to 2."},{"Start":"03:30.440 ","End":"03:34.780","Text":"And we also changed our total amount of oxygens because we still have 2 from the oxygen,"},{"Start":"03:34.780 ","End":"03:37.620","Text":"but we have 2 from the nitric oxide."},{"Start":"03:37.620 ","End":"03:39.660","Text":"We have a total of 4 oxygens."},{"Start":"03:39.660 ","End":"03:43.340","Text":"We\u0027re going to change oxygens to 4 and in our product side,"},{"Start":"03:43.340 ","End":"03:46.100","Text":"we can also see that the nitrogen change to 2,"},{"Start":"03:46.100 ","End":"03:49.190","Text":"2 nitrogens, and our oxygen is 2 times 2,"},{"Start":"03:49.190 ","End":"03:50.390","Text":"so it\u0027s also 4."},{"Start":"03:50.390 ","End":"03:52.160","Text":"You\u0027re going to change the oxygens to 4."},{"Start":"03:52.160 ","End":"03:55.250","Text":"Now we can see that in the reactant side and in"},{"Start":"03:55.250 ","End":"03:58.760","Text":"the product side we have 2 nitrogens and 4 oxygens."},{"Start":"03:58.760 ","End":"04:01.870","Text":"That is our balanced equation for b."},{"Start":"04:01.870 ","End":"04:07.079","Text":"Now we\u0027re going on to c. Nitrate plus"},{"Start":"04:07.079 ","End":"04:12.260","Text":"potassium chromate gives us silver"},{"Start":"04:12.260 ","End":"04:18.800","Text":"chromate plus potassium nitrate."},{"Start":"04:18.800 ","End":"04:23.270","Text":"We\u0027re going to divide this into the reactants and the products and then the reactants,"},{"Start":"04:23.270 ","End":"04:25.840","Text":"we can see we have 1 silver,"},{"Start":"04:25.840 ","End":"04:28.770","Text":"we have 1 nitrate."},{"Start":"04:28.770 ","End":"04:32.840","Text":"Now, remember that the polyatomic ions are taken as a unit."},{"Start":"04:32.840 ","End":"04:39.180","Text":"We have 2 potassium and you have 1 chromate ion,"},{"Start":"04:39.180 ","End":"04:41.075","Text":"again, a polyatomic ion,"},{"Start":"04:41.075 ","End":"04:43.385","Text":"which is taken as a unit."},{"Start":"04:43.385 ","End":"04:46.950","Text":"In our product, we have 2 silver,"},{"Start":"04:47.080 ","End":"04:57.300","Text":"1 nitrate, 1 potassium, and 1 chromate."},{"Start":"05:00.880 ","End":"05:05.210","Text":"First, we can see that the silvers are not balanced."},{"Start":"05:05.210 ","End":"05:10.170","Text":"We have 2 silver in the product and 1 in our reactant."},{"Start":"05:10.170 ","End":"05:13.730","Text":"We\u0027re going to do is we\u0027re going to multiply our silver nitrate by 2."},{"Start":"05:13.730 ","End":"05:17.480","Text":"That way we get 2 silver."},{"Start":"05:17.480 ","End":"05:19.670","Text":"But we also have to change our nitrate."},{"Start":"05:19.670 ","End":"05:22.905","Text":"Now we have 2 so we have 2 nitrates."},{"Start":"05:22.905 ","End":"05:27.145","Text":"Now we can compare our reactants and product size or silver is balanced."},{"Start":"05:27.145 ","End":"05:29.380","Text":"However, in our nitrate is not balanced."},{"Start":"05:29.380 ","End":"05:30.910","Text":"We can see we have 2 on the reactant side,"},{"Start":"05:30.910 ","End":"05:33.145","Text":"and 1 in the product side."},{"Start":"05:33.145 ","End":"05:34.585","Text":"In order to change this,"},{"Start":"05:34.585 ","End":"05:37.910","Text":"we can just multiply our potassium nitrate by 2."},{"Start":"05:38.160 ","End":"05:43.000","Text":"This will change our nitrate to 2,"},{"Start":"05:43.000 ","End":"05:45.790","Text":"and it will also change our potassium to 2."},{"Start":"05:45.790 ","End":"05:50.035","Text":"Now if we compare, we can see that the silver is balanced and nitrate is balanced,"},{"Start":"05:50.035 ","End":"05:53.200","Text":"our potassium is balanced and our chromate is balanced."},{"Start":"05:53.200 ","End":"05:58.150","Text":"That is our final answer for c. Now we\u0027re going on to d. D we have"},{"Start":"05:58.150 ","End":"06:06.640","Text":"silver sulfate plus barium iodide."},{"Start":"06:08.480 ","End":"06:16.680","Text":"Gives us barium sulfate plus silver iodide."},{"Start":"06:16.680 ","End":"06:21.110","Text":"We\u0027re going to divide this into the reactant side and the product side,"},{"Start":"06:21.110 ","End":"06:24.085","Text":"your acting side we can see we have 2 silver,"},{"Start":"06:24.085 ","End":"06:28.190","Text":"and 1 sulfate ion."},{"Start":"06:28.190 ","End":"06:29.960","Text":"This is a polyatomic ion,"},{"Start":"06:29.960 ","End":"06:33.230","Text":"so we don\u0027t separate it into sulfur and oxygen."},{"Start":"06:33.230 ","End":"06:40.260","Text":"We have 1 barium and we have 2 iodines."},{"Start":"06:40.260 ","End":"06:45.160","Text":"In our product side, we have 1 silver,"},{"Start":"06:45.530 ","End":"06:56.325","Text":"1 sulfate ion, 1 barium, and 1 iodine."},{"Start":"06:56.325 ","End":"06:58.710","Text":"We can see that on our reactant side we have"},{"Start":"06:58.710 ","End":"07:00.975","Text":"2 silver and on our product side we have 1 silver."},{"Start":"07:00.975 ","End":"07:03.805","Text":"We can multiply silver iodide by 2."},{"Start":"07:03.805 ","End":"07:07.755","Text":"This way we\u0027re going to change our silver to 2 instead of 1."},{"Start":"07:07.755 ","End":"07:11.815","Text":"This will also change our iodine into 2 instead of 1."},{"Start":"07:11.815 ","End":"07:14.730","Text":"Now we have 2 silver on each side,"},{"Start":"07:14.730 ","End":"07:18.505","Text":"1 sulfate, 1 barium, and 2 iodine."},{"Start":"07:18.505 ","End":"07:21.020","Text":"We can see that everything is balanced."},{"Start":"07:21.020 ","End":"07:26.220","Text":"This is a reaction. That is our final answer for d. Thank you very much for watching."}],"ID":23792},{"Watched":false,"Name":"Chemical Equations and Stoichiometry","Duration":"11m 49s","ChapterTopicVideoID":16923,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.520","Text":"In a previous video,"},{"Start":"00:02.520 ","End":"00:07.050","Text":"we wrote the chemical equation for the combustion of glucose."},{"Start":"00:07.050 ","End":"00:09.600","Text":"In this video, we\u0027ll see that we can use"},{"Start":"00:09.600 ","End":"00:13.830","Text":"the balanced equation to solve chemical problems."},{"Start":"00:13.830 ","End":"00:17.640","Text":"What does this chemical equation tell us?"},{"Start":"00:17.640 ","End":"00:20.565","Text":"When we burn 1 mole of glucose,"},{"Start":"00:20.565 ","End":"00:21.840","Text":"this should be a 1 here,"},{"Start":"00:21.840 ","End":"00:23.130","Text":"but we don\u0027t write it,"},{"Start":"00:23.130 ","End":"00:25.290","Text":"and 6 moles of oxygen,"},{"Start":"00:25.290 ","End":"00:26.640","Text":"here\u0027s the 6,"},{"Start":"00:26.640 ","End":"00:32.115","Text":"we get 6 moles of carbon dioxide and 6 moles of water."},{"Start":"00:32.115 ","End":"00:35.940","Text":"This is result of chemical bookkeeping."},{"Start":"00:35.940 ","End":"00:43.595","Text":"It tells us how many moles we use up in order to produce a certain number of moles."},{"Start":"00:43.595 ","End":"00:46.370","Text":"It\u0027s called stoichiometry."},{"Start":"00:46.370 ","End":"00:51.290","Text":"Stoichiometry, difficult word to spell."},{"Start":"00:51.290 ","End":"00:57.620","Text":"The chemical equation we had for glucose is a simple example of stoichiometry."},{"Start":"00:57.620 ","End":"01:02.299","Text":"Stoichiometry tells us how to perform"},{"Start":"01:02.299 ","End":"01:07.415","Text":"quantitative calculations based on chemical equations."},{"Start":"01:07.415 ","End":"01:09.230","Text":"In order to do this,"},{"Start":"01:09.230 ","End":"01:11.465","Text":"in order to solve problems,"},{"Start":"01:11.465 ","End":"01:14.975","Text":"we need to write conversion factors based,"},{"Start":"01:14.975 ","End":"01:18.020","Text":"for example, on the combustion of glucose."},{"Start":"01:18.020 ","End":"01:21.040","Text":"Let\u0027s write conversion factors."},{"Start":"01:21.040 ","End":"01:23.100","Text":"From 1 mole of glucose,"},{"Start":"01:23.100 ","End":"01:25.545","Text":"we get 6 moles of carbon dioxide."},{"Start":"01:25.545 ","End":"01:29.950","Text":"We can make that into a ratio and write this is equal to 1,"},{"Start":"01:29.950 ","End":"01:36.755","Text":"1 mole of glucose is equivalent in some respect to 6 moles of carbon dioxide."},{"Start":"01:36.755 ","End":"01:44.485","Text":"Similarly, 1 mole of glucose is equivalent to 6 moles of water."},{"Start":"01:44.485 ","End":"01:48.050","Text":"Burning 1 mole of glucose gives us 6 moles of water."},{"Start":"01:48.050 ","End":"01:52.130","Text":"We can make that into a ratio and write equal to 1."},{"Start":"01:52.130 ","End":"01:58.085","Text":"Similarly, 6 moles of water is equivalent to 6 moles of carbon dioxide."},{"Start":"01:58.085 ","End":"01:59.895","Text":"We can divide the 6,"},{"Start":"01:59.895 ","End":"02:06.250","Text":"and write 1 mole of water is equivalent to 1 mole of carbon dioxide."},{"Start":"02:06.250 ","End":"02:09.050","Text":"All these are conversion factors."},{"Start":"02:09.050 ","End":"02:13.190","Text":"We call them stoichiometric ratios."},{"Start":"02:13.190 ","End":"02:17.585","Text":"We can insert them wherever it is relevant."},{"Start":"02:17.585 ","End":"02:19.624","Text":"Let\u0027s take an example."},{"Start":"02:19.624 ","End":"02:23.390","Text":"How many moles of carbon dioxide are produced when"},{"Start":"02:23.390 ","End":"02:28.655","Text":"1.5 moles of glucose is burnt in excess oxygen."},{"Start":"02:28.655 ","End":"02:30.680","Text":"That means we\u0027ve got so much oxygen,"},{"Start":"02:30.680 ","End":"02:32.830","Text":"we don\u0027t need to worry about it."},{"Start":"02:32.830 ","End":"02:36.910","Text":"The number of moles of carbon dioxide is equal to the number of"},{"Start":"02:36.910 ","End":"02:42.895","Text":"moles of glucose multiplied by our conversion factor,"},{"Start":"02:42.895 ","End":"02:46.090","Text":"our stoichiometric ratio, 6 moles"},{"Start":"02:46.090 ","End":"02:49.630","Text":"of carbon dioxide being equivalent to 1 mole of glucose."},{"Start":"02:49.630 ","End":"02:52.240","Text":"First, let\u0027s multiply the numbers."},{"Start":"02:52.240 ","End":"02:55.930","Text":"We start off with 1.5 moles of glucose,"},{"Start":"02:55.930 ","End":"02:59.610","Text":"here it\u0027s 1.5 and here we have 6 divide by 1,"},{"Start":"02:59.610 ","End":"03:01.055","Text":"so we have 6."},{"Start":"03:01.055 ","End":"03:03.085","Text":"Now we need the units."},{"Start":"03:03.085 ","End":"03:06.670","Text":"We have moles of glucose divided by moles of glucose,"},{"Start":"03:06.670 ","End":"03:08.290","Text":"so we can cancel those,"},{"Start":"03:08.290 ","End":"03:11.690","Text":"and we\u0027re left with moles of carbon dioxide."},{"Start":"03:11.820 ","End":"03:16.060","Text":"Multiply 1.5 times 6 and we get,"},{"Start":"03:16.060 ","End":"03:17.660","Text":"of course 9,"},{"Start":"03:17.660 ","End":"03:20.190","Text":"9 moles of carbon dioxide."},{"Start":"03:20.190 ","End":"03:22.295","Text":"That\u0027s the solution to the problem."},{"Start":"03:22.295 ","End":"03:26.029","Text":"When we burn 1.5 moles of glucose,"},{"Start":"03:26.029 ","End":"03:29.300","Text":"we get 9 moles of carbon dioxide."},{"Start":"03:29.300 ","End":"03:31.880","Text":"That\u0027s a very simple problem."},{"Start":"03:31.880 ","End":"03:37.880","Text":"We can also write it and we\u0027ll use this notation in the rest of the video."},{"Start":"03:37.880 ","End":"03:40.460","Text":"Number of moles of carbon dioxide,"},{"Start":"03:40.460 ","End":"03:47.590","Text":"that\u0027s n for numbers of moles and carbon dioxide indicates of what is equal to 9 moles."},{"Start":"03:47.590 ","End":"03:53.150","Text":"We could also write 9 moles of carbon dioxide if we want it to be very exact."},{"Start":"03:53.150 ","End":"03:59.100","Text":"Often the calculations are more complex than just what we had before."},{"Start":"03:59.100 ","End":"04:05.570","Text":"But the central part always involves stoichiometric conversion factors."},{"Start":"04:05.570 ","End":"04:09.710","Text":"The central part is always converting moles to moles,"},{"Start":"04:09.710 ","End":"04:12.380","Text":"moles of products to moles of reactants,"},{"Start":"04:12.380 ","End":"04:14.705","Text":"moles of reactant to moles of products,"},{"Start":"04:14.705 ","End":"04:16.685","Text":"depending on the question."},{"Start":"04:16.685 ","End":"04:19.760","Text":"Let\u0027s write it out like this."},{"Start":"04:19.760 ","End":"04:22.280","Text":"The information we\u0027re given about the reactants,"},{"Start":"04:22.280 ","End":"04:24.965","Text":"we have to convert to the moles of reactants."},{"Start":"04:24.965 ","End":"04:28.865","Text":"Then the moles of reactants have to be converted to the moles of products."},{"Start":"04:28.865 ","End":"04:31.580","Text":"That\u0027s a central part of the problem."},{"Start":"04:31.580 ","End":"04:34.875","Text":"The moles of products have to be converted to"},{"Start":"04:34.875 ","End":"04:38.915","Text":"the information required about the products."},{"Start":"04:38.915 ","End":"04:40.535","Text":"Let\u0027s take an example."},{"Start":"04:40.535 ","End":"04:48.290","Text":"What volume of water is produced when 270 grams of glucose is burnt in excess oxygen."},{"Start":"04:48.290 ","End":"04:54.815","Text":"Let\u0027s convert the scheme that we had above into something a little bit more precise."},{"Start":"04:54.815 ","End":"04:57.935","Text":"We\u0027re given the mass of glucose."},{"Start":"04:57.935 ","End":"05:03.920","Text":"We want to convert it into the moles of glucose."},{"Start":"05:03.920 ","End":"05:06.890","Text":"This is always the central part of the problem."},{"Start":"05:06.890 ","End":"05:11.045","Text":"Then we have to convert it into the moles of water."},{"Start":"05:11.045 ","End":"05:12.920","Text":"From the moles of water,"},{"Start":"05:12.920 ","End":"05:16.295","Text":"we need to convert it into the mass of water,"},{"Start":"05:16.295 ","End":"05:20.255","Text":"and from the mass of water to the volume of water,"},{"Start":"05:20.255 ","End":"05:21.965","Text":"which is what we\u0027re asked about."},{"Start":"05:21.965 ","End":"05:24.440","Text":"We\u0027re asked what volume of water."},{"Start":"05:24.440 ","End":"05:27.785","Text":"We can label this 1,"},{"Start":"05:27.785 ","End":"05:31.080","Text":"2, 3, 4."},{"Start":"05:31.080 ","End":"05:33.235","Text":"These are the 4 steps."},{"Start":"05:33.235 ","End":"05:37.685","Text":"Before we can start to solve this problem."},{"Start":"05:37.685 ","End":"05:42.500","Text":"We need to recall some things that we learned before."},{"Start":"05:42.500 ","End":"05:47.930","Text":"The first thing we need to remember is how to convert the mass to number of moles."},{"Start":"05:47.930 ","End":"05:49.790","Text":"We need the following equation."},{"Start":"05:49.790 ","End":"05:55.790","Text":"We need the number of moles is equal to the mass divided by the molar mass."},{"Start":"05:55.790 ","End":"05:58.925","Text":"Mw, we\u0027re going to write as molar mass."},{"Start":"05:58.925 ","End":"06:02.355","Text":"I know it doesn\u0027t sound right. It\u0027s Mw."},{"Start":"06:02.355 ","End":"06:07.790","Text":"It\u0027s because once we used to call it molecular weight and the name has stuck."},{"Start":"06:07.790 ","End":"06:10.040","Text":"But it\u0027s n = m,"},{"Start":"06:10.040 ","End":"06:12.560","Text":"the mass divided by the molar mass,"},{"Start":"06:12.560 ","End":"06:15.035","Text":"so Mw is molar mass."},{"Start":"06:15.035 ","End":"06:19.520","Text":"At the end, we need to convert the number of moles to mass."},{"Start":"06:19.520 ","End":"06:22.130","Text":"That\u0027s the opposite of what we\u0027ve just done."},{"Start":"06:22.130 ","End":"06:23.930","Text":"We take this equation,"},{"Start":"06:23.930 ","End":"06:27.100","Text":"n = m divided by Mw."},{"Start":"06:27.100 ","End":"06:28.950","Text":"We want m,"},{"Start":"06:28.950 ","End":"06:37.175","Text":"so we can multiply both sides by Mw and we get the mass is equal to n times Mw."},{"Start":"06:37.175 ","End":"06:42.005","Text":"These are just 2 versions of the same equation."},{"Start":"06:42.005 ","End":"06:43.970","Text":"Now before we can proceed,"},{"Start":"06:43.970 ","End":"06:48.035","Text":"we need to know what the molar masses are and we can easily calculate"},{"Start":"06:48.035 ","End":"06:54.535","Text":"the molar mass of glucose is equal to 180 grams."},{"Start":"06:54.535 ","End":"07:00.040","Text":"The molar mass of water is 18 grams, 18.0 grams."},{"Start":"07:00.040 ","End":"07:01.985","Text":"We saw how to do that before."},{"Start":"07:01.985 ","End":"07:06.135","Text":"We need the molar mass of each atom."},{"Start":"07:06.135 ","End":"07:09.695","Text":"We\u0027ve 6 carbon atoms, 12 hydrogen atoms,"},{"Start":"07:09.695 ","End":"07:14.905","Text":"6 oxygen atoms and we add them up and we get to a 180 grams."},{"Start":"07:14.905 ","End":"07:17.000","Text":"In addition, at the end,"},{"Start":"07:17.000 ","End":"07:19.665","Text":"we need to convert the mass to the volume,"},{"Start":"07:19.665 ","End":"07:22.355","Text":"and the mass of water to the volume of water."},{"Start":"07:22.355 ","End":"07:23.870","Text":"How do we do that?"},{"Start":"07:23.870 ","End":"07:26.195","Text":"We learnt this right at the beginning of the course."},{"Start":"07:26.195 ","End":"07:28.190","Text":"We use the equation for density,"},{"Start":"07:28.190 ","End":"07:31.055","Text":"that\u0027s d = m/v."},{"Start":"07:31.055 ","End":"07:34.955","Text":"The density is equal to the mass divided by the volume."},{"Start":"07:34.955 ","End":"07:37.040","Text":"But we need the volume,"},{"Start":"07:37.040 ","End":"07:39.995","Text":"so we multiply both sides by v,"},{"Start":"07:39.995 ","End":"07:43.775","Text":"we get vd = m,"},{"Start":"07:43.775 ","End":"07:52.440","Text":"and from that we get v = m/d by dividing by d. Divide both sides by d,"},{"Start":"07:52.440 ","End":"07:55.380","Text":"we get v = m/d."},{"Start":"07:55.380 ","End":"07:58.465","Text":"This is the equation we require."},{"Start":"07:58.465 ","End":"08:02.675","Text":"d of course is the density and v is the volume."},{"Start":"08:02.675 ","End":"08:05.190","Text":"We need to know what the density of water is"},{"Start":"08:05.190 ","End":"08:07.910","Text":"and this is something we should perhaps remember,"},{"Start":"08:07.910 ","End":"08:12.640","Text":"that the density of water is 1 gram per milliliter."},{"Start":"08:12.640 ","End":"08:18.755","Text":"The density of water is 1 gram per milliliter or 1 kilogram per liter."},{"Start":"08:18.755 ","End":"08:22.610","Text":"In other words, if you carry a 1 liter bottle of water,"},{"Start":"08:22.610 ","End":"08:26.285","Text":"you are carrying 1 kilogram of water."},{"Start":"08:26.285 ","End":"08:29.030","Text":"Let\u0027s start to solve the problem."},{"Start":"08:29.030 ","End":"08:34.130","Text":"The first step, the step 1 is the mass of glucose to the moles of glucose."},{"Start":"08:34.130 ","End":"08:39.004","Text":"We have to convert the mass of glucose to the moles of glucose."},{"Start":"08:39.004 ","End":"08:43.100","Text":"The number of moles of glucose is 270 grams."},{"Start":"08:43.100 ","End":"08:47.490","Text":"That\u0027s the mass of glucose divided by its molar mass,"},{"Start":"08:47.490 ","End":"08:49.925","Text":"180 grams per mole."},{"Start":"08:49.925 ","End":"08:51.435","Text":"We divide the 2,"},{"Start":"08:51.435 ","End":"08:54.405","Text":"we get 1.5 moles."},{"Start":"08:54.405 ","End":"08:58.910","Text":"We have 1.5 moles of glucose."},{"Start":"08:58.910 ","End":"09:00.680","Text":"Now we have Step 2,"},{"Start":"09:00.680 ","End":"09:02.390","Text":"which is the central step."},{"Start":"09:02.390 ","End":"09:06.935","Text":"We have to convert the moles of glucose to moles of water."},{"Start":"09:06.935 ","End":"09:09.635","Text":"The number of moles of water,"},{"Start":"09:09.635 ","End":"09:11.815","Text":"that\u0027s what we need to find out."},{"Start":"09:11.815 ","End":"09:13.670","Text":"After writing moles of water,"},{"Start":"09:13.670 ","End":"09:15.290","Text":"you don\u0027t really need to write it,"},{"Start":"09:15.290 ","End":"09:17.540","Text":"but I want to be very clear,"},{"Start":"09:17.540 ","End":"09:18.995","Text":"at least at the beginning."},{"Start":"09:18.995 ","End":"09:22.490","Text":"Number of moles of water and the units are, of course,"},{"Start":"09:22.490 ","End":"09:28.335","Text":"moles of water is equal the number of moles of glucose,"},{"Start":"09:28.335 ","End":"09:30.595","Text":"and I\u0027ve written moles of glucose,"},{"Start":"09:30.595 ","End":"09:33.125","Text":"times our conversion factor,"},{"Start":"09:33.125 ","End":"09:36.035","Text":"the one that relates water to glucose,"},{"Start":"09:36.035 ","End":"09:40.160","Text":"so 6 moles of water is equivalent to 1 mole of glucose."},{"Start":"09:40.160 ","End":"09:44.410","Text":"Here it is. Let\u0027s solve the numbers first."},{"Start":"09:44.410 ","End":"09:49.365","Text":"We know we have 1.5 moles of glucose,"},{"Start":"09:49.365 ","End":"09:51.750","Text":"we can write 1.5,"},{"Start":"09:51.750 ","End":"09:55.405","Text":"and here we have 6 divided by 1, so 6."},{"Start":"09:55.405 ","End":"09:56.840","Text":"What about the units?"},{"Start":"09:56.840 ","End":"09:58.490","Text":"We have moles of glucose on the top,"},{"Start":"09:58.490 ","End":"10:00.895","Text":"moles of glucose in the bottom."},{"Start":"10:00.895 ","End":"10:04.215","Text":"We\u0027re left with moles of water,"},{"Start":"10:04.215 ","End":"10:08.250","Text":"1.5 times 6 is 9."},{"Start":"10:08.250 ","End":"10:12.420","Text":"Here\u0027s our answer, 9 moles of water."},{"Start":"10:12.420 ","End":"10:14.725","Text":"Now here\u0027s Step 3."},{"Start":"10:14.725 ","End":"10:18.295","Text":"We need to convert the moles of water to the mass of water."},{"Start":"10:18.295 ","End":"10:24.155","Text":"Our equation is, the mass is equal to the number of moles times the molar mass."},{"Start":"10:24.155 ","End":"10:26.285","Text":"We have 9 moles of water."},{"Start":"10:26.285 ","End":"10:30.400","Text":"The molar mass of water is 18.0 grams per mole."},{"Start":"10:30.400 ","End":"10:35.685","Text":"We multiply 9 times 18, we get 162."},{"Start":"10:35.685 ","End":"10:38.970","Text":"Moles cancels with mole minus 1."},{"Start":"10:38.970 ","End":"10:41.940","Text":"Moles times moles minus 1 is just 1."},{"Start":"10:41.940 ","End":"10:44.114","Text":"We\u0027re left just with a gram."},{"Start":"10:44.114 ","End":"10:47.510","Text":"We have a 162 grams."},{"Start":"10:47.510 ","End":"10:50.584","Text":"Now we have the mass of the water produced."},{"Start":"10:50.584 ","End":"10:54.505","Text":"Here we have the final step, Step 4."},{"Start":"10:54.505 ","End":"10:58.040","Text":"The mass of water to the volume of water."},{"Start":"10:58.040 ","End":"11:03.185","Text":"We know from above that the volume\u0027s equal to the mass divide by density."},{"Start":"11:03.185 ","End":"11:05.540","Text":"The mass is 162 grams,"},{"Start":"11:05.540 ","End":"11:07.070","Text":"as we saw up here,"},{"Start":"11:07.070 ","End":"11:08.900","Text":"divided by the density,"},{"Start":"11:08.900 ","End":"11:11.665","Text":"which is 1 gram per milliliter,"},{"Start":"11:11.665 ","End":"11:15.780","Text":"so 162 divided by 1 is just a 162."},{"Start":"11:15.780 ","End":"11:22.725","Text":"Grams cancels with grams and we\u0027re leftover 1 over milliliter to minus 1,"},{"Start":"11:22.725 ","End":"11:24.745","Text":"which is just milliliter."},{"Start":"11:24.745 ","End":"11:30.470","Text":"Here we have our final answer, 162 milliliters."},{"Start":"11:30.470 ","End":"11:32.645","Text":"That\u0027s the answer to the problem."},{"Start":"11:32.645 ","End":"11:35.690","Text":"In this video, we learned about the connection"},{"Start":"11:35.690 ","End":"11:39.005","Text":"between chemical equations and stoichiometry."},{"Start":"11:39.005 ","End":"11:44.245","Text":"We learned how to use stoichiometry to solve numerical problems."},{"Start":"11:44.245 ","End":"11:49.350","Text":"There will be many more examples in the exercises."}],"ID":21133},{"Watched":false,"Name":"Exercise 3","Duration":"1m 30s","ChapterTopicVideoID":20325,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.610 ","End":"00:05.130","Text":"iron metal reacts with chlorine gas."},{"Start":"00:05.130 ","End":"00:08.070","Text":"How many moles of iron chloride are obtained when"},{"Start":"00:08.070 ","End":"00:12.225","Text":"6.78 moles of chlorine reacts with an excess iron."},{"Start":"00:12.225 ","End":"00:16.155","Text":"2 iron plus 3 chlorine gives us 2 iron chloride."},{"Start":"00:16.155 ","End":"00:18.840","Text":"Now if we look at our equation for every 3 moles of chlorine,"},{"Start":"00:18.840 ","End":"00:21.885","Text":"which react, we get 2 moles of iron chloride."},{"Start":"00:21.885 ","End":"00:27.180","Text":"The moles of iron chloride equal the moles of"},{"Start":"00:27.180 ","End":"00:33.730","Text":"chlorine times 2 moles of iron chloride,"},{"Start":"00:36.680 ","End":"00:42.160","Text":"which are achieved for every 3 moles of chlorine which react."},{"Start":"00:44.630 ","End":"00:47.330","Text":"Notice that 3 moles of chlorine are in"},{"Start":"00:47.330 ","End":"00:50.165","Text":"the denominator since we\u0027re multiplying by moles of chlorine,"},{"Start":"00:50.165 ","End":"00:53.840","Text":"that way they will cancel out and we will be left with moles of iron chloride."},{"Start":"00:53.840 ","End":"00:55.790","Text":"This equals moles of chlorine,"},{"Start":"00:55.790 ","End":"01:01.880","Text":"we said is 6.78 moles 6.78 moles of"},{"Start":"01:01.880 ","End":"01:12.850","Text":"chlorine times 2 moles of iron chloride for every 3 moles of chlorine."},{"Start":"01:13.850 ","End":"01:22.450","Text":"This comes to 4.52 moles of iron chloride."},{"Start":"01:23.630 ","End":"01:28.235","Text":"4.52 moles of iron chloride is our final answer."},{"Start":"01:28.235 ","End":"01:31.140","Text":"Thank you very much for watching."}],"ID":21134},{"Watched":false,"Name":"Exercise 4 - Part a","Duration":"","ChapterTopicVideoID":31663,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33900},{"Watched":false,"Name":"Exercise 4 - Part b","Duration":"","ChapterTopicVideoID":31664,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33901},{"Watched":false,"Name":"Exercise 5","Duration":"5m 59s","ChapterTopicVideoID":20327,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.820","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.820 ","End":"00:05.850","Text":"How many grams of silver carbonate are decomposed to"},{"Start":"00:05.850 ","End":"00:10.485","Text":"yield 83.4 grams of silver in the following reaction."},{"Start":"00:10.485 ","End":"00:14.100","Text":"Silver carbonate in heat gives us"},{"Start":"00:14.100 ","End":"00:19.770","Text":"silver plus carbon dioxide plus oxygen and the reaction is not balanced."},{"Start":"00:19.770 ","End":"00:22.125","Text":"First of all, we must balance the reaction."},{"Start":"00:22.125 ","End":"00:31.875","Text":"Silver carbonate gives us silver plus carbon dioxide plus oxygen."},{"Start":"00:31.875 ","End":"00:33.945","Text":"Now if we look at our reactant,"},{"Start":"00:33.945 ","End":"00:38.535","Text":"we have 2 silver here and here we will have 1 silver in our products."},{"Start":"00:38.535 ","End":"00:41.505","Text":"In our reactant, we have 1 carbon."},{"Start":"00:41.505 ","End":"00:43.365","Text":"In the products we have 1 carbon,"},{"Start":"00:43.365 ","End":"00:48.255","Text":"and we have 3 oxygens in our reactants and in our product we have 4 oxygens,"},{"Start":"00:48.255 ","End":"00:51.660","Text":"2 from the oxygen and 2 from the carbon dioxide."},{"Start":"00:51.660 ","End":"00:53.480","Text":"First of all, we\u0027re going to balance our silver"},{"Start":"00:53.480 ","End":"00:55.630","Text":"because here we have 2 and here you have 1."},{"Start":"00:55.630 ","End":"00:57.630","Text":"You can just multiply this by 2."},{"Start":"00:57.630 ","End":"01:01.625","Text":"Next, as we said, we have 1 carbon and on the other side also 1 carbon."},{"Start":"01:01.625 ","End":"01:04.355","Text":"We don\u0027t have to balance the carbons."},{"Start":"01:04.355 ","End":"01:08.750","Text":"We have 3 oxygens here and on this side we have 4 oxygens,"},{"Start":"01:08.750 ","End":"01:11.870","Text":"2 from the carbon dioxide and 2 from the oxygen."},{"Start":"01:11.870 ","End":"01:15.745","Text":"All we have to do here is multiply the oxygen by 1/2."},{"Start":"01:15.745 ","End":"01:21.440","Text":"That way we will have 1 oxygen from here and 2 oxygen from the carbon dioxide equaling 3."},{"Start":"01:21.440 ","End":"01:23.780","Text":"We have 2 silver on both sides,"},{"Start":"01:23.780 ","End":"01:25.745","Text":"1 carbon on both sides,"},{"Start":"01:25.745 ","End":"01:27.485","Text":"and 3 oxygens on both sides."},{"Start":"01:27.485 ","End":"01:30.080","Text":"Now our equation is balanced and we can go on."},{"Start":"01:30.080 ","End":"01:35.090","Text":"Here we\u0027re asked to find the grams of the silver carbonate which"},{"Start":"01:35.090 ","End":"01:40.100","Text":"are decomposed to yield 83.4 grams of silver."},{"Start":"01:40.100 ","End":"01:41.989","Text":"To solve the exercise,"},{"Start":"01:41.989 ","End":"01:44.495","Text":"we\u0027re first going to find the moles of silver"},{"Start":"01:44.495 ","End":"01:47.615","Text":"which were yielded from the mess in the question."},{"Start":"01:47.615 ","End":"01:49.080","Text":"Soon we will you do this."},{"Start":"01:49.080 ","End":"01:51.575","Text":"After we will know the number of moles of silver,"},{"Start":"01:51.575 ","End":"01:55.205","Text":"we will calculate the number of moles of silver carbonate,"},{"Start":"01:55.205 ","End":"01:58.490","Text":"and then we\u0027ll convert these into the mass of the sulfur carbonate."},{"Start":"01:58.490 ","End":"02:02.135","Text":"First of all, we want to calculate the number of moles of the silver."},{"Start":"02:02.135 ","End":"02:07.410","Text":"So n moles equals m mass divided by Mw,"},{"Start":"02:07.410 ","End":"02:08.745","Text":"which is molar mass."},{"Start":"02:08.745 ","End":"02:15.560","Text":"Again, the moles of the silver Ag equals the mass of the silver,"},{"Start":"02:15.560 ","End":"02:17.780","Text":"which equals 83.4 grams."},{"Start":"02:17.780 ","End":"02:20.150","Text":"The mass of the silver,"},{"Start":"02:20.150 ","End":"02:24.020","Text":"divided by the molar mass of silver and this equals"},{"Start":"02:24.020 ","End":"02:28.190","Text":"83.4 grams divided by the molar mass of silver,"},{"Start":"02:28.190 ","End":"02:32.835","Text":"which equals 107.87 grams per mole."},{"Start":"02:32.835 ","End":"02:36.760","Text":"This is taken from the periodic table of elements."},{"Start":"02:36.760 ","End":"02:42.360","Text":"This comes to 0.77 moles of silver."},{"Start":"02:42.360 ","End":"02:46.850","Text":"Now the next step is to find the moles of the silver carbonate."},{"Start":"02:46.850 ","End":"02:52.220","Text":"Moles of silver carbonate equal"},{"Start":"02:52.220 ","End":"02:57.365","Text":"the number of moles of silver times if we look at our reaction,"},{"Start":"02:57.365 ","End":"03:00.950","Text":"we can see that for every 1 mole of silver carbonate which reacts,"},{"Start":"03:00.950 ","End":"03:03.095","Text":"we get 2 moles of silver."},{"Start":"03:03.095 ","End":"03:12.590","Text":"It\u0027s times 1 mole of silver carbonate divided by 2 moles of silver."},{"Start":"03:12.590 ","End":"03:15.380","Text":"The moles of silver are in the denominator,"},{"Start":"03:15.380 ","End":"03:17.540","Text":"because we want the moles of silver to cancel out."},{"Start":"03:17.540 ","End":"03:24.470","Text":"This equals again the moles of silver here we calculated them 0.77 mole of"},{"Start":"03:24.470 ","End":"03:29.570","Text":"silver times 1 mole of"},{"Start":"03:29.570 ","End":"03:36.105","Text":"silver carbonate divided by 2 moles of silver."},{"Start":"03:36.105 ","End":"03:44.820","Text":"The moles of silver cancel out and we get 0.39 mole silver carbonate."},{"Start":"03:45.460 ","End":"03:50.450","Text":"Now the next step is to calculate the mass of the silver carbonate."},{"Start":"03:50.450 ","End":"03:52.670","Text":"In order to find the mass of the silver carbonate,"},{"Start":"03:52.670 ","End":"03:59.885","Text":"we will use the equation n moles equals MS divided by the molar mass as we used before."},{"Start":"03:59.885 ","End":"04:02.470","Text":"Now we\u0027re interested in calculating the mass."},{"Start":"04:02.470 ","End":"04:07.570","Text":"The mass of the silver carbonate equals,"},{"Start":"04:07.570 ","End":"04:10.805","Text":"if we multiply both sides by the molar mass,"},{"Start":"04:10.805 ","End":"04:13.505","Text":"the moles times the molar mass."},{"Start":"04:13.505 ","End":"04:19.205","Text":"Moles, of course, of silver carbonate times the molar mass of silver carbonate."},{"Start":"04:19.205 ","End":"04:22.145","Text":"We already calculated the moles of silver carbonate."},{"Start":"04:22.145 ","End":"04:23.435","Text":"We have the answer here."},{"Start":"04:23.435 ","End":"04:25.490","Text":"Now we have to calculate the molar mass."},{"Start":"04:25.490 ","End":"04:33.440","Text":"The molar mass of silver carbonate equals 2 times the molar mass of silver,"},{"Start":"04:33.440 ","End":"04:36.155","Text":"since we have 2 silvers in silver carbonate,"},{"Start":"04:36.155 ","End":"04:38.240","Text":"plus the molar mass of carbon."},{"Start":"04:38.240 ","End":"04:40.355","Text":"Since we have 1 carbon in silver carbonate,"},{"Start":"04:40.355 ","End":"04:43.115","Text":"plus 3 times the molar mass of oxygen."},{"Start":"04:43.115 ","End":"04:45.680","Text":"Since we have 3 oxygens in silver carbonate."},{"Start":"04:45.680 ","End":"04:51.060","Text":"This equals 2 times 107.87 grams per mole plus"},{"Start":"04:51.060 ","End":"04:58.780","Text":"12.01 grams per mole plus 3 times 16 grams per mole."},{"Start":"04:58.780 ","End":"05:02.575","Text":"The molar masses were taken from the periodic table of elements."},{"Start":"05:02.575 ","End":"05:08.475","Text":"This equals to 75.75 grams per mole."},{"Start":"05:08.475 ","End":"05:12.035","Text":"That is the molar mass of the silver carbonate."},{"Start":"05:12.035 ","End":"05:14.750","Text":"Now we can go back to our equation."},{"Start":"05:14.750 ","End":"05:20.390","Text":"The mass of the silver carbonate equals"},{"Start":"05:20.390 ","End":"05:26.850","Text":"the moles of silver carbonate times the molar mass of silver carbonate."},{"Start":"05:28.880 ","End":"05:32.600","Text":"This equals the moles we calculated before is equal"},{"Start":"05:32.600 ","End":"05:35.975","Text":"0.39 mole times the molar mass of silver carbonate,"},{"Start":"05:35.975 ","End":"05:42.690","Text":"which we also calculated and equals 275.75 grams per mole."},{"Start":"05:43.150 ","End":"05:45.605","Text":"The moles cancel out,"},{"Start":"05:45.605 ","End":"05:51.410","Text":"and this equals 107.54 grams."},{"Start":"05:51.410 ","End":"05:55.850","Text":"The mass of the silver carbonate equals 107.54 grams."},{"Start":"05:55.850 ","End":"05:57.500","Text":"That is our final answer."},{"Start":"05:57.500 ","End":"06:00.060","Text":"Thank you very much for watching."}],"ID":21136},{"Watched":false,"Name":"Exercise 6a","Duration":"4m 23s","ChapterTopicVideoID":20330,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.310","Text":"Hey, we\u0027re going to solve the following exercise."},{"Start":"00:02.310 ","End":"00:06.345","Text":"Calcium hydride plus water gives us calcium hydroxide plus hydrogen,"},{"Start":"00:06.345 ","End":"00:07.950","Text":"and the equation is not balanced."},{"Start":"00:07.950 ","End":"00:11.490","Text":"A, how many moles of hydrogen result from the reaction"},{"Start":"00:11.490 ","End":"00:15.420","Text":"of 130 grams calcium hydride with an excess of water?"},{"Start":"00:15.420 ","End":"00:18.150","Text":"B, how many grams of water are consumed in"},{"Start":"00:18.150 ","End":"00:21.907","Text":"the reaction of 65.2 grams of calcium hydride?"},{"Start":"00:21.907 ","End":"00:26.285","Text":"C, what mass of calcium hydride must react with an excess of water"},{"Start":"00:26.285 ","End":"00:31.370","Text":"to produce 6.5 times 10^24 molecules of hydrogen?"},{"Start":"00:31.370 ","End":"00:33.680","Text":"Before we begin solving a, b,"},{"Start":"00:33.680 ","End":"00:37.085","Text":"c, we\u0027re going to start by balancing the equation."},{"Start":"00:37.085 ","End":"00:42.350","Text":"We have calcium hydride plus water gives"},{"Start":"00:42.350 ","End":"00:48.475","Text":"us calcium hydroxide plus hydrogen."},{"Start":"00:48.475 ","End":"00:52.355","Text":"We have 1 calcium on the reactant side."},{"Start":"00:52.355 ","End":"00:58.514","Text":"We also have 4 hydrogens and we have 1 oxygen."},{"Start":"00:58.514 ","End":"01:01.740","Text":"On the product side, we have 1 calcium,"},{"Start":"01:01.740 ","End":"01:04.440","Text":"we have 4 hydrogens,"},{"Start":"01:04.440 ","End":"01:08.245","Text":"and we have 2 oxygens because it\u0027s O_2 here."},{"Start":"01:08.245 ","End":"01:10.680","Text":"If we look at the calciums, they\u0027re balanced."},{"Start":"01:10.680 ","End":"01:12.200","Text":"If we look at the hydrogens, they\u0027re also balanced."},{"Start":"01:12.200 ","End":"01:15.025","Text":"If we look at the oxygens on the reactant side we have 1,"},{"Start":"01:15.025 ","End":"01:16.760","Text":"on the product side, we have 2."},{"Start":"01:16.760 ","End":"01:20.150","Text":"In order to change this oxygen into 2 on the reactant side,"},{"Start":"01:20.150 ","End":"01:22.060","Text":"we\u0027re going to multiply the water by 2,"},{"Start":"01:22.060 ","End":"01:24.655","Text":"that way our oxygen will change to 2."},{"Start":"01:24.655 ","End":"01:26.870","Text":"We\u0027re also changing our hydrogens,"},{"Start":"01:26.870 ","End":"01:28.970","Text":"the hydrogen number will change to 6."},{"Start":"01:28.970 ","End":"01:34.040","Text":"Now again, we look at the calcium and they are balanced and oxygens are also balanced."},{"Start":"01:34.040 ","End":"01:36.440","Text":"The hydrogens are the only ones that are not balanced."},{"Start":"01:36.440 ","End":"01:38.330","Text":"In our reactant side, we have 6,"},{"Start":"01:38.330 ","End":"01:40.360","Text":"in our product side, we have 4."},{"Start":"01:40.360 ","End":"01:42.360","Text":"We want 6 hydrogens on"},{"Start":"01:42.360 ","End":"01:46.385","Text":"our product side and the easiest way to do this is with the hydrogen."},{"Start":"01:46.385 ","End":"01:48.260","Text":"If we multiply our hydrogen by 2,"},{"Start":"01:48.260 ","End":"01:50.270","Text":"we\u0027re going to have 4 hydrogens from"},{"Start":"01:50.270 ","End":"01:54.555","Text":"the hydrogen and 2 more hydrogens from the calcium hydroxide,"},{"Start":"01:54.555 ","End":"01:56.955","Text":"so we have 6 hydrogens."},{"Start":"01:56.955 ","End":"01:59.235","Text":"Now we have 1 calcium on each side,"},{"Start":"01:59.235 ","End":"02:01.815","Text":"6 hydrogens, and 2 oxygens."},{"Start":"02:01.815 ","End":"02:04.580","Text":"Now the equation is balanced and we can go on to a."},{"Start":"02:04.580 ","End":"02:09.365","Text":"Again, in a we\u0027re asked to find the number of moles of the hydrogen."},{"Start":"02:09.365 ","End":"02:14.375","Text":"In order to do this, we\u0027ll first calculate the moles of the calcium hydride,"},{"Start":"02:14.375 ","End":"02:17.830","Text":"then calculate the moles of the hydrogen. Let\u0027s begin."},{"Start":"02:17.830 ","End":"02:20.450","Text":"To calculate the moles of the calcium hydride we\u0027re going to use"},{"Start":"02:20.450 ","End":"02:23.790","Text":"the equation n number of moles equals m,"},{"Start":"02:23.790 ","End":"02:26.250","Text":"the mass, divided by the molar mass."},{"Start":"02:26.250 ","End":"02:31.290","Text":"The mass is given n=130 grams and the molar mass we have to calculate."},{"Start":"02:31.290 ","End":"02:36.790","Text":"The molar mass of calcium hydride equals the molar mass of"},{"Start":"02:36.790 ","End":"02:44.020","Text":"calcium plus 2 times the molar mass of hydrogen since we have 1 calcium, 2 hydrogens."},{"Start":"02:44.020 ","End":"02:54.970","Text":"This equals 40.01 grams per mole plus 2 times 1.01 grams per mole,"},{"Start":"02:54.970 ","End":"02:59.990","Text":"which equals 42.03 grams per mole."},{"Start":"03:00.090 ","End":"03:03.190","Text":"That\u0027s the molar mass of the calcium hydride."},{"Start":"03:03.190 ","End":"03:06.325","Text":"Now we\u0027re going to calculate the number of moles of calcium hydride."},{"Start":"03:06.325 ","End":"03:07.720","Text":"The number of moles n of"},{"Start":"03:07.720 ","End":"03:11.605","Text":"the calcium hydride equals the mass of the calcium hydride from here,"},{"Start":"03:11.605 ","End":"03:14.485","Text":"divided by the molar mass of the calcium hydride."},{"Start":"03:14.485 ","End":"03:19.910","Text":"This equals 130 grams, which is given,"},{"Start":"03:19.910 ","End":"03:25.775","Text":"divided by 42.03 grams per mole and this equals 3.09 mole."},{"Start":"03:25.775 ","End":"03:28.985","Text":"The number of moles of the calcium hydride are 3.09."},{"Start":"03:28.985 ","End":"03:33.050","Text":"The next step is to calculate the number of moles of the hydrogen."},{"Start":"03:33.050 ","End":"03:38.400","Text":"The moles of the hydrogen equal the moles of calcium hydride."},{"Start":"03:38.400 ","End":"03:40.430","Text":"If we look at our reaction,"},{"Start":"03:40.430 ","End":"03:43.025","Text":"for every 1 mole of calcium hydride which reacts,"},{"Start":"03:43.025 ","End":"03:44.975","Text":"we get 2 moles of hydrogen."},{"Start":"03:44.975 ","End":"03:48.800","Text":"We\u0027re going to multiply the moles of calcium hydride by"},{"Start":"03:48.800 ","End":"03:53.130","Text":"2 moles of hydrogen for every 1 mole of calcium hydride."},{"Start":"03:53.130 ","End":"03:58.485","Text":"This equals the moles of calcium hydride which we calculated and equal 3.09 mole,"},{"Start":"03:58.485 ","End":"04:05.850","Text":"times 2 moles of hydrogen for every 1 mole of calcium hydride."},{"Start":"04:05.850 ","End":"04:10.280","Text":"The calcium hydride cancels out,"},{"Start":"04:10.280 ","End":"04:13.520","Text":"and we get 6.19 mole."},{"Start":"04:13.520 ","End":"04:17.270","Text":"The moles of hydrogen equals 6.19 mole."},{"Start":"04:17.270 ","End":"04:18.980","Text":"That is our final answer for a,"},{"Start":"04:18.980 ","End":"04:20.405","Text":"thank you very much for watching."},{"Start":"04:20.405 ","End":"04:23.220","Text":"We will continue to b in the next video."}],"ID":21139},{"Watched":false,"Name":"Exercise 6b","Duration":"3m 37s","ChapterTopicVideoID":20328,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.730","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.730 ","End":"00:04.875","Text":"In the last video, we solved a,"},{"Start":"00:04.875 ","End":"00:06.060","Text":"now we\u0027re going on to b."},{"Start":"00:06.060 ","End":"00:11.745","Text":"How many grams of water are consumed in the reaction of 65.2 grams of calcium hydride."},{"Start":"00:11.745 ","End":"00:17.010","Text":"Calcium hydride plus 2 water gives us calcium hydroxide plus hydrogen,"},{"Start":"00:17.010 ","End":"00:18.660","Text":"the equation was balanced in a."},{"Start":"00:18.660 ","End":"00:20.190","Text":"Now we\u0027re going onto b."},{"Start":"00:20.190 ","End":"00:21.810","Text":"In order to answer b,"},{"Start":"00:21.810 ","End":"00:25.160","Text":"we will first calculate the number of moles of calcium hydride,"},{"Start":"00:25.160 ","End":"00:26.915","Text":"then the number of moles of water,"},{"Start":"00:26.915 ","End":"00:29.710","Text":"and convert the number of moles to grams."},{"Start":"00:29.710 ","End":"00:34.640","Text":"Remember that the number of moles equals mass n divided by molar mass."},{"Start":"00:34.640 ","End":"00:39.695","Text":"In a, we calculated the molar mass of calcium hydride."},{"Start":"00:39.695 ","End":"00:43.610","Text":"It equals 42.03 grams per mole."},{"Start":"00:43.610 ","End":"00:48.170","Text":"The number of moles of calcium hydride equal the mass of"},{"Start":"00:48.170 ","End":"00:53.435","Text":"the calcium hydride divided by the molar mass of calcium hydride."},{"Start":"00:53.435 ","End":"00:55.550","Text":"This equals 65.2 grams,"},{"Start":"00:55.550 ","End":"00:57.005","Text":"which is given in the question,"},{"Start":"00:57.005 ","End":"00:58.880","Text":"divided by our molar mass,"},{"Start":"00:58.880 ","End":"01:02.135","Text":"which is 42.03, which was calculated in a,"},{"Start":"01:02.135 ","End":"01:03.790","Text":"grams per mole, of course,"},{"Start":"01:03.790 ","End":"01:06.599","Text":"and this equals 1.55 mole."},{"Start":"01:06.599 ","End":"01:08.300","Text":"Now, I just want to remind you,"},{"Start":"01:08.300 ","End":"01:10.970","Text":"when you\u0027re dividing by a fraction, for example,"},{"Start":"01:10.970 ","End":"01:12.845","Text":"grams divided by grams per mole,"},{"Start":"01:12.845 ","End":"01:16.220","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction,"},{"Start":"01:16.220 ","End":"01:19.250","Text":"meaning it\u0027s the same as grams times mole per grams."},{"Start":"01:19.250 ","End":"01:21.260","Text":"The grams cancel out, and we\u0027re left with moles."},{"Start":"01:21.260 ","End":"01:25.460","Text":"Now we know the number of moles of calcium hydride equal 1.55 mole."},{"Start":"01:25.460 ","End":"01:28.145","Text":"Now we\u0027re going to calculate the moles of the water."},{"Start":"01:28.145 ","End":"01:32.360","Text":"The moles of the water equal the moles of calcium hydride,"},{"Start":"01:32.360 ","End":"01:35.090","Text":"if we look at our equation,"},{"Start":"01:35.090 ","End":"01:38.075","Text":"we can see that for every 1 mole calcium hydride which reacts,"},{"Start":"01:38.075 ","End":"01:40.310","Text":"we have 2 moles of water which are reacting,"},{"Start":"01:40.310 ","End":"01:45.780","Text":"it\u0027s times 2 moles of water divided by 1 mole of calcium hydride."},{"Start":"01:46.310 ","End":"01:49.145","Text":"Now the moles of calcium hydride already in"},{"Start":"01:49.145 ","End":"01:51.340","Text":"the denominator that way they will cancel out,"},{"Start":"01:51.340 ","End":"01:52.975","Text":"and we\u0027ll get the moles of the water."},{"Start":"01:52.975 ","End":"01:56.215","Text":"This equals 1.55 mole, which we calculated."},{"Start":"01:56.215 ","End":"02:03.390","Text":"Calcium hydride times 2 moles of water divided by 1 mole calcium hydroxide."},{"Start":"02:03.390 ","End":"02:06.535","Text":"The moles of calcium hydroxide cancel out,"},{"Start":"02:06.535 ","End":"02:09.200","Text":"and we get 3.1 mole."},{"Start":"02:09.890 ","End":"02:12.950","Text":"The moles of the water are 3.1."},{"Start":"02:12.950 ","End":"02:15.400","Text":"Now all we have to do is convert these moles into grams."},{"Start":"02:15.400 ","End":"02:17.155","Text":"We\u0027ll use the equation we used before."},{"Start":"02:17.155 ","End":"02:20.140","Text":"Number of moles equals mass divided by molar mass."},{"Start":"02:20.140 ","End":"02:22.090","Text":"We\u0027re looking for the mass."},{"Start":"02:22.090 ","End":"02:27.370","Text":"Therefore, the mass of the water equals the number of"},{"Start":"02:27.370 ","End":"02:32.715","Text":"moles of the water times the molar mass of the water,"},{"Start":"02:32.715 ","End":"02:35.390","Text":"we\u0027re just multiplying both sides by the molar mass."},{"Start":"02:35.390 ","End":"02:38.563","Text":"The moles we calculated,"},{"Start":"02:38.563 ","End":"02:39.710","Text":"and we have 3.1 mole,"},{"Start":"02:39.710 ","End":"02:41.555","Text":"and we have to calculate the molar mass of water."},{"Start":"02:41.555 ","End":"02:45.890","Text":"The molar mass of water equals the molar mass of"},{"Start":"02:45.890 ","End":"02:51.530","Text":"hydrogen times 2 since we have 2 hydrogens plus the molar mass of oxygen."},{"Start":"02:51.530 ","End":"03:00.650","Text":"This equals 2 times 1.01 grams per mole plus 16 grams per mole."},{"Start":"03:00.650 ","End":"03:04.495","Text":"This equals 18.02 grams per mole."},{"Start":"03:04.495 ","End":"03:07.580","Text":"The molar masses of the hydrogen and oxygen,"},{"Start":"03:07.580 ","End":"03:10.190","Text":"we\u0027re taking from the periodic table of elements."},{"Start":"03:10.190 ","End":"03:13.880","Text":"Now that we have the molar mass of the water, we can continue."},{"Start":"03:13.880 ","End":"03:15.665","Text":"The number of moles of the water,"},{"Start":"03:15.665 ","End":"03:19.640","Text":"which we calculated are 3.1 mole times the molar mass of the water,"},{"Start":"03:19.640 ","End":"03:21.980","Text":"which is 18.02 grams per mole,"},{"Start":"03:21.980 ","End":"03:28.165","Text":"the moles cancel out and this equals 55.87 grams."},{"Start":"03:28.165 ","End":"03:32.325","Text":"The mass of the water equals 55.87 grams."},{"Start":"03:32.325 ","End":"03:34.090","Text":"That is our final answer for b."},{"Start":"03:34.090 ","End":"03:35.660","Text":"Thank you very much for watching."},{"Start":"03:35.660 ","End":"03:38.520","Text":"We\u0027ll continue to c in the next video."}],"ID":21137},{"Watched":false,"Name":"Exercise 6c","Duration":"3m 39s","ChapterTopicVideoID":20329,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.145 ","End":"00:04.890","Text":"In previous videos, we solved a and b."},{"Start":"00:04.890 ","End":"00:09.690","Text":"Now we\u0027re going down to c. What mass of calcium hydride must react with an excess"},{"Start":"00:09.690 ","End":"00:14.880","Text":"of water to produce 6.5 times 10^24 molecules of hydrogen?"},{"Start":"00:14.880 ","End":"00:17.835","Text":"From the number of molecules of hydrogen and which are given."},{"Start":"00:17.835 ","End":"00:20.735","Text":"We\u0027re going to calculate the number of moles of hydrogen,"},{"Start":"00:20.735 ","End":"00:24.283","Text":"then calculate the number of moles of the calcium hydride,"},{"Start":"00:24.283 ","End":"00:26.390","Text":"and then convert to the mass of the calcium hydride."},{"Start":"00:26.390 ","End":"00:29.299","Text":"So to calculate the number of moles of the hydrogen,"},{"Start":"00:29.299 ","End":"00:36.590","Text":"we take the number of molecules of hydrogen divided by the Avogadro number."},{"Start":"00:36.590 ","End":"00:41.330","Text":"Because the Avogadro number states how many molecules we have in each mole."},{"Start":"00:41.330 ","End":"00:48.845","Text":"So the number of molecules of the hydrogen equals 6.5 times 10^24 molecules."},{"Start":"00:48.845 ","End":"00:52.820","Text":"And this is divided by the Avogadro number,"},{"Start":"00:52.820 ","End":"00:55.310","Text":"which states that there are 6.02 times"},{"Start":"00:55.310 ","End":"01:03.395","Text":"10^23 molecules in every mole"},{"Start":"01:03.395 ","End":"01:08.440","Text":"and this equals 10.45 mole."},{"Start":"01:08.440 ","End":"01:10.510","Text":"I\u0027m just reminding you again,"},{"Start":"01:10.510 ","End":"01:12.265","Text":"when we divide by a fraction,"},{"Start":"01:12.265 ","End":"01:14.750","Text":"which here we have molecules,"},{"Start":"01:15.260 ","End":"01:19.120","Text":"divided by molecules per mole."},{"Start":"01:21.990 ","End":"01:26.140","Text":"It\u0027s the same as multiplying by the reciprocal."},{"Start":"01:26.140 ","End":"01:31.460","Text":"So that\u0027s molecules times mole per molecules."},{"Start":"01:31.560 ","End":"01:35.155","Text":"The molecules cancel out, and we\u0027re left with moles."},{"Start":"01:35.155 ","End":"01:39.490","Text":"So now we know that the moles of the hydrogen are 10.45 mole."},{"Start":"01:39.490 ","End":"01:43.000","Text":"The next step is to find the moles of the calcium hydride."},{"Start":"01:43.000 ","End":"01:50.075","Text":"So the number of moles of calcium hydride equal the number of moles of hydrogen times."},{"Start":"01:50.075 ","End":"01:54.800","Text":"If we look at our equation for every 1 mole of calcium hydride which reacts,"},{"Start":"01:54.800 ","End":"01:56.955","Text":"we get 2 moles of hydrogen."},{"Start":"01:56.955 ","End":"02:03.645","Text":"So it\u0027s times 1 mole of calcium hydroxide for every 2 moles of hydrogen."},{"Start":"02:03.645 ","End":"02:05.660","Text":"Moles of hydrogen are in the denominator."},{"Start":"02:05.660 ","End":"02:06.845","Text":"That way they cancel out,"},{"Start":"02:06.845 ","End":"02:08.900","Text":"and we\u0027re left with moles of calcium hydride."},{"Start":"02:08.900 ","End":"02:14.750","Text":"So this equals 10.45 mole of hydrogen times 1"},{"Start":"02:14.750 ","End":"02:21.244","Text":"mol calcium hydride divided by 2 moles of hydrogen."},{"Start":"02:21.244 ","End":"02:30.250","Text":"The moles of hydrogen cancel out and we get 5.23 moles of calcium hydride."},{"Start":"02:30.250 ","End":"02:32.800","Text":"So our next step is to convert these moles into"},{"Start":"02:32.800 ","End":"02:35.695","Text":"grams in order to get the mass of the calcium hydride."},{"Start":"02:35.695 ","End":"02:37.645","Text":"So for this purpose, we\u0027re going to use the equation,"},{"Start":"02:37.645 ","End":"02:41.220","Text":"n number of moles equals m mass divided by MW,"},{"Start":"02:41.220 ","End":"02:42.510","Text":"which is the molar mass."},{"Start":"02:42.510 ","End":"02:44.815","Text":"We want to find the mass,"},{"Start":"02:44.815 ","End":"02:49.210","Text":"the mass of the calcium hydride equals the number of moles"},{"Start":"02:49.210 ","End":"02:54.070","Text":"of calcium hydride times the molar mass of calcium hydride."},{"Start":"02:54.070 ","End":"02:58.790","Text":"So the number of moles we calculated here, 5.23."},{"Start":"03:00.290 ","End":"03:05.140","Text":"The molar mass of the calcium hydride was calculated in a."},{"Start":"03:05.140 ","End":"03:13.700","Text":"The molar mass of calcium hydride equals 42.03 g/mol."},{"Start":"03:13.700 ","End":"03:21.875","Text":"So this equals 5.23 mol times 42.03 g/mol."},{"Start":"03:21.875 ","End":"03:24.530","Text":"Moles cancel out."},{"Start":"03:24.530 ","End":"03:29.870","Text":"We get 219.82 graphs."},{"Start":"03:29.870 ","End":"03:34.280","Text":"So the mass of the calcium hydride equals 219.82"},{"Start":"03:34.280 ","End":"03:39.930","Text":"g. That is our final answer for C. Thank you very much for watching."}],"ID":21138},{"Watched":false,"Name":"Limiting Reactants","Duration":"10m 49s","ChapterTopicVideoID":16924,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.530","Text":"In a previous video,"},{"Start":"00:02.530 ","End":"00:06.330","Text":"we balanced the chemical equation for the combustion of glucose."},{"Start":"00:06.330 ","End":"00:11.010","Text":"We found that 1 mole of glucose reacts with 6 moles of oxygen"},{"Start":"00:11.010 ","End":"00:16.409","Text":"to give 6 moles of carbon dioxide and 6 moles of water."},{"Start":"00:16.409 ","End":"00:21.300","Text":"Now we can make this equation a little bit more precise by adding some notation."},{"Start":"00:21.300 ","End":"00:24.705","Text":"We can add s for solid, l for liquid,"},{"Start":"00:24.705 ","End":"00:28.995","Text":"g for gas, and aq for if it\u0027s dissolved in water."},{"Start":"00:28.995 ","End":"00:32.265","Text":"That\u0027s from Aqua which is a Latin for water."},{"Start":"00:32.265 ","End":"00:38.060","Text":"What we get is glucose which is a solid reacting with oxygen which is"},{"Start":"00:38.060 ","End":"00:43.940","Text":"a gas to give carbon dioxide which is a gas and water which is a liquid."},{"Start":"00:43.940 ","End":"00:48.800","Text":"Or if it\u0027s very high temperature as it probably is when you burn something,"},{"Start":"00:48.800 ","End":"00:50.995","Text":"it could be a gas."},{"Start":"00:50.995 ","End":"00:56.690","Text":"Often we write the conditions under which the reaction is performed above the arrow."},{"Start":"00:56.690 ","End":"01:01.280","Text":"For example, this Delta is capital"},{"Start":"01:01.280 ","End":"01:06.620","Text":"Delta means that it is performed at very high temperatures."},{"Start":"01:06.620 ","End":"01:11.530","Text":"Now in a previous video we also talked about stoichiometry."},{"Start":"01:11.530 ","End":"01:14.925","Text":"Remember chemical bookkeeping."},{"Start":"01:14.925 ","End":"01:23.110","Text":"These numbers, the number of moles tell us precisely how much of each component we need."},{"Start":"01:23.110 ","End":"01:26.935","Text":"1 mole of glucose will react with 6 moles of"},{"Start":"01:26.935 ","End":"01:32.800","Text":"oxygen and give us 6 moles of carbon dioxide and 6 moles of water."},{"Start":"01:32.800 ","End":"01:36.010","Text":"Now sometimes we don\u0027t have enough of one of"},{"Start":"01:36.010 ","End":"01:39.160","Text":"the reactants for the reaction to go to completion."},{"Start":"01:39.160 ","End":"01:43.825","Text":"That means that some of the reactants are left when the reaction is finished."},{"Start":"01:43.825 ","End":"01:46.690","Text":"Now the substance, the component,"},{"Start":"01:46.690 ","End":"01:50.260","Text":"the reactant that prevents the reaction from going to"},{"Start":"01:50.260 ","End":"01:54.100","Text":"completion is called a limiting reactant."},{"Start":"01:54.100 ","End":"01:55.915","Text":"Let\u0027s take some examples."},{"Start":"01:55.915 ","End":"02:00.559","Text":"Example 1, supposing we want to burn 1 mole of glucose,"},{"Start":"02:00.559 ","End":"02:04.100","Text":"but only have 5 moles of oxygen available,"},{"Start":"02:04.100 ","End":"02:08.905","Text":"how much glucose will remain after all the oxygen is consumed?"},{"Start":"02:08.905 ","End":"02:11.840","Text":"Now when we wrote the chemical equation,"},{"Start":"02:11.840 ","End":"02:17.105","Text":"we saw that we need 6 moles of oxygen to completely burn 1 mole of glucose."},{"Start":"02:17.105 ","End":"02:19.365","Text":"This problem we already have 5."},{"Start":"02:19.365 ","End":"02:23.930","Text":"Obviously we don\u0027t have enough to burn all the glucose."},{"Start":"02:23.930 ","End":"02:26.615","Text":"How much glucose will we actually burn?"},{"Start":"02:26.615 ","End":"02:30.880","Text":"We first need the conversion fact or the stoichiometric ratio."},{"Start":"02:30.880 ","End":"02:35.285","Text":"That\u0027s 1 mole of glucose is equivalent to 6 moles of oxygen."},{"Start":"02:35.285 ","End":"02:39.785","Text":"1 mole of glucose requires 6 moles of oxygen to burn."},{"Start":"02:39.785 ","End":"02:43.195","Text":"We can write this ratio and it\u0027s equal to 1."},{"Start":"02:43.195 ","End":"02:45.680","Text":"That\u0027s our stoichiometric ratio."},{"Start":"02:45.680 ","End":"02:47.280","Text":"Now here\u0027s the calculation."},{"Start":"02:47.280 ","End":"02:49.980","Text":"The number of moles of glucose."},{"Start":"02:49.980 ","End":"02:54.210","Text":"I\u0027ve written out in total and full."},{"Start":"02:54.210 ","End":"02:56.284","Text":"Number of moles of glucose,"},{"Start":"02:56.284 ","End":"02:59.869","Text":"moles of glucose is equal to the number of moles of oxygen,"},{"Start":"02:59.869 ","End":"03:02.420","Text":"moles of oxygen that\u0027s units."},{"Start":"03:02.420 ","End":"03:05.760","Text":"Here\u0027s our stoichiometric ratio."},{"Start":"03:05.760 ","End":"03:10.250","Text":"1 mole of glucose divided by 6 moles of oxygen."},{"Start":"03:10.250 ","End":"03:13.880","Text":"Now we can put in the numbers and oxygen,"},{"Start":"03:13.880 ","End":"03:16.630","Text":"number of moles of oxygen is 5."},{"Start":"03:16.630 ","End":"03:22.769","Text":"Here\u0027s the 6, so it\u0027s 5 divided by 6 5/6ths and we need the units."},{"Start":"03:22.769 ","End":"03:26.040","Text":"We have moles of oxygen on the numerator,"},{"Start":"03:26.040 ","End":"03:28.665","Text":"moles of oxygen in the denominator,"},{"Start":"03:28.665 ","End":"03:31.440","Text":"and we\u0027re left with moles of glucose."},{"Start":"03:31.440 ","End":"03:34.905","Text":"We have 5/6 moles of glucose."},{"Start":"03:34.905 ","End":"03:39.800","Text":"From that, we can conclude that only 5/6 moles of"},{"Start":"03:39.800 ","End":"03:44.930","Text":"glucose will be used up and 1/6 of the mole will be left."},{"Start":"03:44.930 ","End":"03:47.965","Text":"Because we started off with 1 whole mole."},{"Start":"03:47.965 ","End":"03:49.965","Text":"We used 5/6th,"},{"Start":"03:49.965 ","End":"03:52.115","Text":"so 1/6th will be left."},{"Start":"03:52.115 ","End":"03:54.515","Text":"What could we say about that?"},{"Start":"03:54.515 ","End":"03:58.385","Text":"We can say that oxygen is the limiting reactant."},{"Start":"03:58.385 ","End":"04:03.265","Text":"Oxygen is what prevented us from burning all the glucose."},{"Start":"04:03.265 ","End":"04:05.615","Text":"Since we had glucose leftover,"},{"Start":"04:05.615 ","End":"04:08.405","Text":"glucose is in excess."},{"Start":"04:08.405 ","End":"04:10.820","Text":"Let\u0027s take a more complicated example."},{"Start":"04:10.820 ","End":"04:12.230","Text":"Example 2."},{"Start":"04:12.230 ","End":"04:16.190","Text":"If 88.3 grams of sulfuric acid reacts with"},{"Start":"04:16.190 ","End":"04:22.385","Text":"68.0 grams of sodium hydroxide to form sodium sulfate and water,"},{"Start":"04:22.385 ","End":"04:24.995","Text":"which is the limiting reactant?"},{"Start":"04:24.995 ","End":"04:28.774","Text":"The first thing we need to do is to write the equation."},{"Start":"04:28.774 ","End":"04:31.535","Text":"Now what reaction is this?"},{"Start":"04:31.535 ","End":"04:34.340","Text":"This reaction is a reaction between"},{"Start":"04:34.340 ","End":"04:40.365","Text":"a strong acid and a strong base sodium hydroxide base."},{"Start":"04:40.365 ","End":"04:44.825","Text":"It\u0027s going to give us a salt plus water."},{"Start":"04:44.825 ","End":"04:49.280","Text":"The first step is to write the equation and balance it."},{"Start":"04:49.280 ","End":"04:51.005","Text":"I\u0027m going to write the equation."},{"Start":"04:51.005 ","End":"04:54.110","Text":"You don\u0027t yet know how to write such an equation."},{"Start":"04:54.110 ","End":"04:55.840","Text":"I\u0027m going to write it for you."},{"Start":"04:55.840 ","End":"04:58.540","Text":"We\u0027ll learn more about it later."},{"Start":"04:58.540 ","End":"05:03.800","Text":"Here\u0027s sulfuric acid H_2SO_4 and it\u0027s in water,"},{"Start":"05:03.800 ","End":"05:07.370","Text":"so it\u0027s aqueous plus NaOH,"},{"Start":"05:07.370 ","End":"05:11.765","Text":"that\u0027s sodium hydroxide and it\u0027s also in water aq,"},{"Start":"05:11.765 ","End":"05:15.065","Text":"to give sodium sulfate."},{"Start":"05:15.065 ","End":"05:18.500","Text":"That\u0027s also in water because it\u0027s soluble in water."},{"Start":"05:18.500 ","End":"05:23.735","Text":"Here\u0027s the water. We get a little bit extra water, L for liquid."},{"Start":"05:23.735 ","End":"05:26.890","Text":"Now we need to balance this equation."},{"Start":"05:26.890 ","End":"05:30.270","Text":"If we look at the equation we see"},{"Start":"05:30.270 ","End":"05:38.460","Text":"that there\u0027s only 1 sodium on the left hand side and 2 on the right hand side."},{"Start":"05:38.460 ","End":"05:42.130","Text":"We need to multiply Na by 2,"},{"Start":"05:42.130 ","End":"05:46.435","Text":"we can only multiply the whole of NaOH."},{"Start":"05:46.435 ","End":"05:48.615","Text":"Let\u0027s look at the hydrogens."},{"Start":"05:48.615 ","End":"05:50.460","Text":"2 hydrogens here."},{"Start":"05:50.460 ","End":"05:53.565","Text":"Another 2 hydrogens here 2 times 1."},{"Start":"05:53.565 ","End":"05:55.685","Text":"That\u0027s 4 altogether."},{"Start":"05:55.685 ","End":"06:01.920","Text":"Here we have just 2 H_2O, just 2 hydrogens."},{"Start":"06:01.920 ","End":"06:06.525","Text":"We need to multiply water by 2 to make 4 hydrogens."},{"Start":"06:06.525 ","End":"06:09.416","Text":"Now the equation is balanced."},{"Start":"06:09.416 ","End":"06:18.340","Text":"H_2SO_4 plus 2NaOH to give Na_2SO_4 plus 2H2O."},{"Start":"06:18.340 ","End":"06:20.951","Text":"Now we can write the number of moles."},{"Start":"06:20.951 ","End":"06:24.740","Text":"1 mole of sulfuric acid,"},{"Start":"06:24.740 ","End":"06:31.760","Text":"2 moles of sodium hydroxide to give 1 mole of sodium sulfate and 2 moles of water."},{"Start":"06:31.760 ","End":"06:34.940","Text":"The second step is to calculate the number of moles"},{"Start":"06:34.940 ","End":"06:38.555","Text":"of each substance using the molar masses."},{"Start":"06:38.555 ","End":"06:45.295","Text":"The molar mass of sulfuric acid we can work it out is 98.1 grams per mole."},{"Start":"06:45.295 ","End":"06:49.985","Text":"The molar mass of sodium hydroxide is 40.0 grams per mole."},{"Start":"06:49.985 ","End":"06:53.330","Text":"Now the number of moles of sulfuric acid is the mass of"},{"Start":"06:53.330 ","End":"06:58.165","Text":"sulfuric acid which is 88.3 grams we\u0027re given that in the question,"},{"Start":"06:58.165 ","End":"07:03.570","Text":"divided by the molar mass 98.1 grams per mole,"},{"Start":"07:03.570 ","End":"07:07.230","Text":"that gives us 0.90 moles."},{"Start":"07:07.230 ","End":"07:14.240","Text":"The number of moles of sodium hydroxide is the mass of sodium hydroxide 68.0 grams"},{"Start":"07:14.240 ","End":"07:22.015","Text":"divided by the molar mass 40.0 grams per mole and that gives us 1.7 moles."},{"Start":"07:22.015 ","End":"07:26.780","Text":"Step 3 is to check which reactant is the limiting one."},{"Start":"07:26.780 ","End":"07:33.440","Text":"Let\u0027s look at the situation that we have 0.90"},{"Start":"07:33.440 ","End":"07:39.935","Text":"moles of sulfuric acid and see how much sodium hydroxide reacts."},{"Start":"07:39.935 ","End":"07:47.840","Text":"Or we could have 1.7 moles of sodium hydroxide and see how much sulfuric acid reacts."},{"Start":"07:47.840 ","End":"07:51.080","Text":"To do this, we first need the conversion factor"},{"Start":"07:51.080 ","End":"07:54.200","Text":"for the reactants from the chemical equation."},{"Start":"07:54.200 ","End":"07:59.780","Text":"It\u0027s 1 mole sulfuric acid to 2 moles of sodium hydroxide."},{"Start":"07:59.780 ","End":"08:01.680","Text":"That\u0027s our conversion factor."},{"Start":"08:01.680 ","End":"08:09.590","Text":"Now if the number of moles of sulfuric acid is 0.90 moles then the number of moles of"},{"Start":"08:09.590 ","End":"08:13.400","Text":"sodium hydroxide we require is 0.90"},{"Start":"08:13.400 ","End":"08:18.145","Text":"moles of sulfuric acid multiplied by the conversion factor,"},{"Start":"08:18.145 ","End":"08:22.940","Text":"2 moles of sodium hydroxide to 1 mole of sulfuric acid."},{"Start":"08:22.940 ","End":"08:25.100","Text":"We can work out the numbers first,"},{"Start":"08:25.100 ","End":"08:29.900","Text":"0.90 times 2 gives us 1.8."},{"Start":"08:29.900 ","End":"08:36.215","Text":"Then the units moles of sulfuric acid cancels with moles of sulfuric acid,"},{"Start":"08:36.215 ","End":"08:40.190","Text":"leaves us moles of sodium hydroxide and the answer we"},{"Start":"08:40.190 ","End":"08:45.060","Text":"obtain is 1.8 moles of sodium hydroxide."},{"Start":"08:45.060 ","End":"08:49.120","Text":"Now let\u0027s consider the second possibility."},{"Start":"08:50.750 ","End":"08:55.824","Text":"Supposing we have 1.7 moles of sodium hydroxide,"},{"Start":"08:55.824 ","End":"08:59.950","Text":"then how many moles of sulfuric acid will react?"},{"Start":"08:59.950 ","End":"09:03.530","Text":"The moles of sulfuric acid is 1.7 moles of"},{"Start":"09:03.530 ","End":"09:07.396","Text":"sodium hydroxide multiplied by our conversion factor,"},{"Start":"09:07.396 ","End":"09:12.244","Text":"1 mole of sulfuric acid to 2 moles of sodium hydroxide."},{"Start":"09:12.244 ","End":"09:19.770","Text":"Let\u0027s take the numbers 1.7 divided by 2 is 0.85 and then the units."},{"Start":"09:19.770 ","End":"09:23.900","Text":"Moles of sodium hydroxide cancels with moles of"},{"Start":"09:23.900 ","End":"09:28.750","Text":"sodium hydroxide to give us moles of sulfuric acid."},{"Start":"09:28.750 ","End":"09:33.050","Text":"We have 0.85 moles of sulfuric acid."},{"Start":"09:33.050 ","End":"09:40.925","Text":"Now for all the sulfuric acid to react we need 1.8 moles of sodium hydroxide."},{"Start":"09:40.925 ","End":"09:42.470","Text":"That\u0027s what we have here."},{"Start":"09:42.470 ","End":"09:45.740","Text":"But we only have 1.7 moles."},{"Start":"09:45.740 ","End":"09:47.825","Text":"That\u0027s what we calculated before."},{"Start":"09:47.825 ","End":"09:50.120","Text":"We only have 1.7."},{"Start":"09:50.120 ","End":"09:55.355","Text":"We don\u0027t have enough for all the sulfuric acid to react."},{"Start":"09:55.355 ","End":"10:00.455","Text":"Therefore sodium hydroxide is the limiting reactant."},{"Start":"10:00.455 ","End":"10:02.900","Text":"We can look at it another way."},{"Start":"10:02.900 ","End":"10:11.530","Text":"We need 0.85 moles of sulfuric acid for all the NaOH to react."},{"Start":"10:11.530 ","End":"10:14.535","Text":"We have 0.9 moles."},{"Start":"10:14.535 ","End":"10:18.155","Text":"There\u0027ll be sulfuric acid left after the reaction."},{"Start":"10:18.155 ","End":"10:22.590","Text":"It will be in excess. We have 0.9."},{"Start":"10:22.590 ","End":"10:25.860","Text":"We only require 0.85."},{"Start":"10:25.860 ","End":"10:32.715","Text":"We can calculate how much will be left over 0.05 moles."},{"Start":"10:32.715 ","End":"10:36.320","Text":"In this video we saw that if the reactants are not in"},{"Start":"10:36.320 ","End":"10:42.005","Text":"the stoichiometric ratio that was defined by the chemical equation,"},{"Start":"10:42.005 ","End":"10:49.020","Text":"one reactant will be the limiting reactant and the other will be in excess."}],"ID":21141},{"Watched":false,"Name":"Stoichiometry in Solutions","Duration":"13m 31s","ChapterTopicVideoID":16925,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.565","Text":"In a previous video,"},{"Start":"00:02.565 ","End":"00:05.910","Text":"we introduced the concept of stoichiometry."},{"Start":"00:05.910 ","End":"00:10.620","Text":"In this video, we will discuss stoichiometry in solutions."},{"Start":"00:10.620 ","End":"00:11.940","Text":"In order to begin,"},{"Start":"00:11.940 ","End":"00:13.785","Text":"we need some definitions."},{"Start":"00:13.785 ","End":"00:15.915","Text":"We need to define solute,"},{"Start":"00:15.915 ","End":"00:17.880","Text":"solvent, and solution."},{"Start":"00:17.880 ","End":"00:21.375","Text":"Let\u0027s begin with solvent."},{"Start":"00:21.375 ","End":"00:26.415","Text":"The solvent is the substance in which the substance of interest is dissolved."},{"Start":"00:26.415 ","End":"00:28.170","Text":"It could be, for example,"},{"Start":"00:28.170 ","End":"00:31.215","Text":"as it will be in the rest of this video, water."},{"Start":"00:31.215 ","End":"00:37.020","Text":"Or it could be some organic solvent used to clean your clothes."},{"Start":"00:37.020 ","End":"00:40.655","Text":"The solute is a substance dissolved in the solvent."},{"Start":"00:40.655 ","End":"00:45.080","Text":"For example, if I dissolve salt in water,"},{"Start":"00:45.080 ","End":"00:47.090","Text":"salt will be the solute,"},{"Start":"00:47.090 ","End":"00:49.430","Text":"and water will be the solvent."},{"Start":"00:49.430 ","End":"00:53.750","Text":"The solution is obtained when the solute is dissolved in the solvent."},{"Start":"00:53.750 ","End":"00:59.045","Text":"For example, salty water is a solution of salt in water."},{"Start":"00:59.045 ","End":"01:03.395","Text":"In this video, we will talk mainly about aqueous solutions,"},{"Start":"01:03.395 ","End":"01:07.415","Text":"solutions in water, but there are many other solutions."},{"Start":"01:07.415 ","End":"01:13.700","Text":"For example, air is a solution where the solvent is nitrogen,"},{"Start":"01:13.700 ","End":"01:18.250","Text":"and the solute is oxygen,"},{"Start":"01:18.250 ","End":"01:21.785","Text":"or carbon dioxide, or a few other things."},{"Start":"01:21.785 ","End":"01:24.905","Text":"Or another example is soda."},{"Start":"01:24.905 ","End":"01:26.960","Text":"Soda, which we drink."},{"Start":"01:26.960 ","End":"01:32.090","Text":"Soda is a solution of carbon dioxide in water."},{"Start":"01:32.090 ","End":"01:35.420","Text":"Now let\u0027s talk about aqueous solutions."},{"Start":"01:35.420 ","End":"01:40.060","Text":"We\u0027re going to take as an example a solution of sodium chloride in water."},{"Start":"01:40.060 ","End":"01:45.235","Text":"Here sodium chloride is the solute and water is the solvent,"},{"Start":"01:45.235 ","End":"01:50.495","Text":"we write it as NaCl with aq in brackets after it,"},{"Start":"01:50.495 ","End":"01:53.885","Text":"that\u0027s indicating an aqueous solution."},{"Start":"01:53.885 ","End":"01:56.780","Text":"Now we\u0027re going to talk about concentration."},{"Start":"01:56.780 ","End":"02:00.845","Text":"We can have solutions of varying concentrations."},{"Start":"02:00.845 ","End":"02:04.010","Text":"We could have a very salty solution that would be"},{"Start":"02:04.010 ","End":"02:08.495","Text":"a higher concentration than the barely salty solution."},{"Start":"02:08.495 ","End":"02:14.555","Text":"Now there are many ways of defining concentration."},{"Start":"02:14.555 ","End":"02:19.175","Text":"The one we\u0027re going to talk about this video is called molarity."},{"Start":"02:19.175 ","End":"02:25.579","Text":"Rewrite it as capital M. The molarity of a solution is defined as follows."},{"Start":"02:25.579 ","End":"02:34.760","Text":"The molarity is equal to n where it\u0027s the number of moles of solute divided by V,"},{"Start":"02:34.760 ","End":"02:38.495","Text":"that\u0027s the volume of the solution in liters."},{"Start":"02:38.495 ","End":"02:42.845","Text":"The volume must be in liters, not in milliliters."},{"Start":"02:42.845 ","End":"02:46.400","Text":"How can we prepare an aqueous solution?"},{"Start":"02:46.400 ","End":"02:50.824","Text":"In this picture, we have a volumetric flask."},{"Start":"02:50.824 ","End":"02:52.745","Text":"This is this flask."},{"Start":"02:52.745 ","End":"02:56.450","Text":"It has a mark on it giving a particular volume,"},{"Start":"02:56.450 ","End":"02:59.270","Text":"here it\u0027s 250 milliliters."},{"Start":"02:59.270 ","End":"03:02.135","Text":"We put the solid we\u0027re interested in,"},{"Start":"03:02.135 ","End":"03:04.640","Text":"in the base of the flask."},{"Start":"03:04.640 ","End":"03:05.990","Text":"Usually we weigh it,"},{"Start":"03:05.990 ","End":"03:07.760","Text":"and then put it there."},{"Start":"03:07.760 ","End":"03:14.270","Text":"Then we add a little of the solvent sufficient to dissolve the solute,"},{"Start":"03:14.270 ","End":"03:16.100","Text":"and then finally,"},{"Start":"03:16.100 ","End":"03:25.235","Text":"we add enough solvent to reach the mark on the neck of the flask."},{"Start":"03:25.235 ","End":"03:29.870","Text":"This flask is called a volumetric flask."},{"Start":"03:29.870 ","End":"03:33.950","Text":"We can find them in all different volumes."},{"Start":"03:33.950 ","End":"03:37.895","Text":"Let\u0027s take an example based on this picture."},{"Start":"03:37.895 ","End":"03:42.260","Text":"What is the molarity of an aqueous solution of NaCl when"},{"Start":"03:42.260 ","End":"03:49.300","Text":"0.584 grams is dissolved in 250 milliliters of water?"},{"Start":"03:49.300 ","End":"03:52.715","Text":"In order to calculate the number of moles,"},{"Start":"03:52.715 ","End":"03:57.410","Text":"we first need to know what the molar mass is of NaCl."},{"Start":"03:57.410 ","End":"04:03.005","Text":"The molar mass of NaCl is 58.4 grams per mole."},{"Start":"04:03.005 ","End":"04:09.470","Text":"We get it by adding up the molar mass of sodium to the molar mass of chlorine."},{"Start":"04:09.470 ","End":"04:14.765","Text":"Now the number of moles of NaCl is its mass divided by molar mass."},{"Start":"04:14.765 ","End":"04:17.885","Text":"It\u0027s mass, according to the question we\u0027ve been asked,"},{"Start":"04:17.885 ","End":"04:20.705","Text":"is 0.584 grams,"},{"Start":"04:20.705 ","End":"04:24.785","Text":"and its molar mass is 58.4 grams per mole."},{"Start":"04:24.785 ","End":"04:26.955","Text":"When we divide these 2,"},{"Start":"04:26.955 ","End":"04:29.030","Text":"let\u0027s deal with the numbers first of all,"},{"Start":"04:29.030 ","End":"04:31.240","Text":"we get 0.01,"},{"Start":"04:31.240 ","End":"04:32.580","Text":"and then the units,"},{"Start":"04:32.580 ","End":"04:35.055","Text":"the grams cancels with grams."},{"Start":"04:35.055 ","End":"04:37.950","Text":"1 over molar to the minus 1 is mole,"},{"Start":"04:37.950 ","End":"04:42.525","Text":"so we get 0.01 moles of NaCl."},{"Start":"04:42.525 ","End":"04:45.135","Text":"Now we need the volume."},{"Start":"04:45.135 ","End":"04:48.710","Text":"We were told that the volume is 250 milliliters,"},{"Start":"04:48.710 ","End":"04:51.745","Text":"but we need the volume in liters."},{"Start":"04:51.745 ","End":"04:55.010","Text":"We multiply it by this conversion factor,"},{"Start":"04:55.010 ","End":"04:59.030","Text":"1 liter is the same as 1,000 milliliters,"},{"Start":"04:59.030 ","End":"05:04.100","Text":"250 divided by 1,000 is 0.254."},{"Start":"05:04.100 ","End":"05:08.920","Text":"For units the milliliters here cancels with milliliters here,"},{"Start":"05:08.920 ","End":"05:11.100","Text":"and we are left with liters,"},{"Start":"05:11.100 ","End":"05:14.960","Text":"so the volume is 0.25 liters."},{"Start":"05:14.960 ","End":"05:17.075","Text":"Now we can calculate the molarity."},{"Start":"05:17.075 ","End":"05:21.155","Text":"Molarity is the number of moles, 0.01 moles,"},{"Start":"05:21.155 ","End":"05:27.605","Text":"divided by the volume of the solution in liters, 0.25 liters."},{"Start":"05:27.605 ","End":"05:31.240","Text":"Divide the numbers we get 0.04."},{"Start":"05:31.240 ","End":"05:34.470","Text":"The units are moles per liter."},{"Start":"05:34.870 ","End":"05:40.960","Text":"We can say the molarity is 0.04M."},{"Start":"05:40.960 ","End":"05:43.275","Text":"That\u0027s the molarity,"},{"Start":"05:43.275 ","End":"05:48.020","Text":"that\u0027s the concentration of the salt in water."},{"Start":"05:48.020 ","End":"05:53.750","Text":"Now sometimes we need to perform the opposite of what we\u0027ve just done."},{"Start":"05:53.750 ","End":"05:59.200","Text":"We need to calculate the number of moles from the volume, and concentration."},{"Start":"05:59.200 ","End":"06:01.720","Text":"Let\u0027s take the equation for the molarity."},{"Start":"06:01.720 ","End":"06:05.110","Text":"The Molarity is number of moles divided by the volume."},{"Start":"06:05.110 ","End":"06:09.100","Text":"If we multiply both sides of this equation by V,"},{"Start":"06:09.100 ","End":"06:13.740","Text":"we get on 1 side m times V,"},{"Start":"06:13.740 ","End":"06:17.640","Text":"that\u0027s here, and the other side we get n,"},{"Start":"06:17.640 ","End":"06:21.135","Text":"we have n equals M times V."},{"Start":"06:21.135 ","End":"06:25.644","Text":"The number of moles is equal to the molarity times the volume."},{"Start":"06:25.644 ","End":"06:27.235","Text":"Let\u0027s take an example."},{"Start":"06:27.235 ","End":"06:30.655","Text":"How many moles of NaCl are required to prepare"},{"Start":"06:30.655 ","End":"06:36.350","Text":"100 milliliters of 1 molar aqueous solution?"},{"Start":"06:36.350 ","End":"06:42.095","Text":"Now, we know that the number of moles is equal to the molarity times the volume."},{"Start":"06:42.095 ","End":"06:44.010","Text":"That\u0027s 1 M,"},{"Start":"06:44.010 ","End":"06:48.965","Text":"that\u0027s the molarity times the volume 0.1 liters."},{"Start":"06:48.965 ","End":"06:52.280","Text":"Now we can write out the units in full,"},{"Start":"06:52.280 ","End":"06:56.405","Text":"M is moles per liter."},{"Start":"06:56.405 ","End":"06:59.750","Text":"We have moles per liter times liter,"},{"Start":"06:59.750 ","End":"07:01.655","Text":"that just leaves moles,"},{"Start":"07:01.655 ","End":"07:08.885","Text":"and here\u0027s the moles, and 1 times 0.1 is of course just 0.1."},{"Start":"07:08.885 ","End":"07:13.414","Text":"The answer is 0.1 moles."},{"Start":"07:13.414 ","End":"07:18.875","Text":"Now we\u0027re going to use what we\u0027ve just calculated in a chemical reaction."},{"Start":"07:18.875 ","End":"07:23.090","Text":"The chemical reaction we have molarity, and stoichiometry."},{"Start":"07:23.090 ","End":"07:25.925","Text":"We\u0027re going to connect all these 3 concepts."},{"Start":"07:25.925 ","End":"07:29.300","Text":"Here\u0027s a reaction we considered in the previous video."},{"Start":"07:29.300 ","End":"07:36.125","Text":"Sulfuric acid reacts with sodium hydroxide to get sodium sulfate, and water."},{"Start":"07:36.125 ","End":"07:38.810","Text":"The reaction is already balanced."},{"Start":"07:38.810 ","End":"07:40.640","Text":"Here are the number of moles."},{"Start":"07:40.640 ","End":"07:43.370","Text":"1 mole of sulfuric acid,"},{"Start":"07:43.370 ","End":"07:49.295","Text":"and 2 moles of sodium hydroxide gives us 1 mole of sodium sulfate,"},{"Start":"07:49.295 ","End":"07:51.070","Text":"and 2 moles of water."},{"Start":"07:51.070 ","End":"07:57.380","Text":"Let\u0027s use an example based on this equation.100 milliliters of 0.1"},{"Start":"07:57.380 ","End":"08:05.230","Text":"molar H_2SO_4 reacts with 200 milliliters of 0.1 molar NaOH,"},{"Start":"08:05.230 ","End":"08:09.325","Text":"what is the molarity of Na_2SO_4?"},{"Start":"08:09.325 ","End":"08:14.285","Text":"We\u0027re given information about the reactants,"},{"Start":"08:14.285 ","End":"08:18.475","Text":"and we\u0027re expected to find information about the products."},{"Start":"08:18.475 ","End":"08:22.295","Text":"Here\u0027s the path we discussed in another video."},{"Start":"08:22.295 ","End":"08:27.605","Text":"The information about the reactants will lead us to calculate the moles of reactants."},{"Start":"08:27.605 ","End":"08:30.559","Text":"From that we\u0027ll calculate the moles of products."},{"Start":"08:30.559 ","End":"08:34.175","Text":"This is the central part of the problem."},{"Start":"08:34.175 ","End":"08:39.955","Text":"From the moles and products we\u0027ll find information about the products."},{"Start":"08:39.955 ","End":"08:43.620","Text":"To relate this a little better to the question we\u0027ve been asked,"},{"Start":"08:43.620 ","End":"08:45.325","Text":"we can write it like this."},{"Start":"08:45.325 ","End":"08:47.920","Text":"The concentration of volume of reactants,"},{"Start":"08:47.920 ","End":"08:50.919","Text":"that\u0027s what\u0027s given us in the question."},{"Start":"08:50.919 ","End":"08:54.400","Text":"From this we\u0027ll calculate the moles of reactants."},{"Start":"08:54.400 ","End":"08:57.670","Text":"From that we\u0027ll go to the moles of product,"},{"Start":"08:57.670 ","End":"09:01.975","Text":"and from that we\u0027ll go to the volume and concentration of the product."},{"Start":"09:01.975 ","End":"09:03.760","Text":"These are the stages,"},{"Start":"09:03.760 ","End":"09:07.445","Text":"1, 2, 3."},{"Start":"09:07.445 ","End":"09:11.355","Text":"There are 3 stages in solving this problem."},{"Start":"09:11.355 ","End":"09:17.015","Text":"In Step 1, we need to calculate the moles of sulfuric acid."},{"Start":"09:17.015 ","End":"09:23.700","Text":"The number of moles of sulfuric acid is equal to its molarity times its volume,"},{"Start":"09:23.700 ","End":"09:31.025","Text":"0.1 M times 0.1 liters."},{"Start":"09:31.025 ","End":"09:38.880","Text":"I\u0027ve already converted the 100 milliliters to 0.1 liters."},{"Start":"09:38.880 ","End":"09:43.790","Text":"If I multiply 0.1 times 0.1, I get 0.01."},{"Start":"09:43.790 ","End":"09:46.745","Text":"If I calculate capital M,"},{"Start":"09:46.745 ","End":"09:53.270","Text":"which you remember is moles per liter by liter,"},{"Start":"09:53.270 ","End":"09:55.150","Text":"I get just moles."},{"Start":"09:55.150 ","End":"09:58.515","Text":"I get 0.01 moles."},{"Start":"09:58.515 ","End":"10:03.665","Text":"Now, the number of moles of sodium hydroxide is again,"},{"Start":"10:03.665 ","End":"10:06.725","Text":"it\u0027s molarity times volume."},{"Start":"10:06.725 ","End":"10:13.010","Text":"We have 0.1 molar times 0.2 liters."},{"Start":"10:13.010 ","End":"10:18.950","Text":"Again, we get 0.02 and the units are moles."},{"Start":"10:18.950 ","End":"10:23.670","Text":"We\u0027re not going to spell it out every time."},{"Start":"10:23.670 ","End":"10:26.420","Text":"M times L,"},{"Start":"10:26.420 ","End":"10:32.374","Text":"molarity times liters will always give us moles."},{"Start":"10:32.374 ","End":"10:37.835","Text":"The moles of sulfuric acid is 0.01,"},{"Start":"10:37.835 ","End":"10:42.770","Text":"and the moles of NaOH is 0.02."},{"Start":"10:42.770 ","End":"10:49.205","Text":"Going to take a little detour to discuss whether there is a limiting reactant."},{"Start":"10:49.205 ","End":"10:52.940","Text":"Let\u0027s write the number of moles of NAOH,"},{"Start":"10:52.940 ","End":"10:58.520","Text":"that\u0027s 0.02 divided by the number of moles of sulfuric acid,"},{"Start":"10:58.520 ","End":"11:07.860","Text":"that 0.01, we divide 0.02 by 0.01 we get 2 divided by 1."},{"Start":"11:07.860 ","End":"11:12.562","Text":"That\u0027s 2 moles of NAOH divided by 1 mole of H_2SO_4,"},{"Start":"11:12.562 ","End":"11:16.955","Text":"and that\u0027s precisely the stoichiometric ratio."},{"Start":"11:16.955 ","End":"11:19.250","Text":"That\u0027s equal to 1."},{"Start":"11:19.250 ","End":"11:22.790","Text":"From all this, we can conclude that all the reactants will"},{"Start":"11:22.790 ","End":"11:27.425","Text":"react completely because they\u0027re in stoichiometric proportions."},{"Start":"11:27.425 ","End":"11:29.150","Text":"Now Step 2,"},{"Start":"11:29.150 ","End":"11:33.700","Text":"how many moles of Na_2SO_4 will be formed?"},{"Start":"11:33.700 ","End":"11:36.120","Text":"You look at the stoichiometric ratio,"},{"Start":"11:36.120 ","End":"11:42.165","Text":"1 mole of Na_2SO_4 is equivalent to 1 mole of sulfuric acid,"},{"Start":"11:42.165 ","End":"11:44.490","Text":"that ratio gives us 1."},{"Start":"11:44.490 ","End":"11:47.750","Text":"The number of moles of Na_2SO_4 will be"},{"Start":"11:47.750 ","End":"11:52.055","Text":"precisely equal to the number of moles of H_2SO_4."},{"Start":"11:52.055 ","End":"11:54.915","Text":"We found that was 0.01."},{"Start":"11:54.915 ","End":"12:00.385","Text":"The number of moles of Na_2SO_4 will be 0.01."},{"Start":"12:00.385 ","End":"12:03.290","Text":"Now, in order to complete the problem,"},{"Start":"12:03.290 ","End":"12:07.760","Text":"we need the volume and concentration of Na_2SO_4."},{"Start":"12:07.760 ","End":"12:09.455","Text":"Now in the beginning,"},{"Start":"12:09.455 ","End":"12:12.335","Text":"we had 100 milliliters of sulfuric acid,"},{"Start":"12:12.335 ","End":"12:14.945","Text":"and 200 milliliters of NAOH,"},{"Start":"12:14.945 ","End":"12:17.935","Text":"the total is 300 milliliters."},{"Start":"12:17.935 ","End":"12:23.840","Text":"The small amount of water formed in the reaction,"},{"Start":"12:23.840 ","End":"12:28.880","Text":"is really quite negligible with respect to 300 milliliters,"},{"Start":"12:28.880 ","End":"12:31.075","Text":"so we\u0027ll just ignore it,"},{"Start":"12:31.075 ","End":"12:33.985","Text":"we have 300 milliliters."},{"Start":"12:33.985 ","End":"12:38.975","Text":"Now we can calculate the molarity because we know the number of moles,"},{"Start":"12:38.975 ","End":"12:40.555","Text":"and we know the volume."},{"Start":"12:40.555 ","End":"12:43.680","Text":"The number of moles is 0.01 moles,"},{"Start":"12:43.680 ","End":"12:48.010","Text":"the volume is 0.30 liters,"},{"Start":"12:48.010 ","End":"12:53.555","Text":"I\u0027ve converted 300 milliliters to 0.30 liters."},{"Start":"12:53.555 ","End":"12:57.765","Text":"If I divide 0.01 by 0.30,"},{"Start":"12:57.765 ","End":"13:03.930","Text":"I get 0.033 in moles per liter. That\u0027s the units."},{"Start":"13:03.930 ","End":"13:07.140","Text":"Here\u0027s the answer."},{"Start":"13:07.140 ","End":"13:13.040","Text":"This tells us that the Na_2SO_4 that\u0027s formed has"},{"Start":"13:13.040 ","End":"13:19.250","Text":"a molarity of 0.033M."},{"Start":"13:19.250 ","End":"13:24.950","Text":"In this video, we"},{"Start":"13:24.950 ","End":"13:31.680","Text":"learned about stoichiometry when the reaction involves aqueous solutions."}],"ID":21142},{"Watched":false,"Name":"Exercise 7","Duration":"4m 40s","ChapterTopicVideoID":20331,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.030 ","End":"00:08.180","Text":"A piece of aluminum foil measuring 10.60 cm times 5.20"},{"Start":"00:08.180 ","End":"00:14.145","Text":"cm times 0.375 mm is dissolved in excess hydrogen chloride."},{"Start":"00:14.145 ","End":"00:16.170","Text":"What mass of hydrogen is produced?"},{"Start":"00:16.170 ","End":"00:20.305","Text":"The density of aluminum is 2.70 gram per centimeters^3."},{"Start":"00:20.305 ","End":"00:22.085","Text":"We have our equation."},{"Start":"00:22.085 ","End":"00:28.560","Text":"2 aluminum plus 6 hydrogen chloride gives us 2 aluminum chloride plus 3 hydrogen."},{"Start":"00:28.560 ","End":"00:30.679","Text":"We need to find the mass of hydrogen."},{"Start":"00:30.679 ","End":"00:32.350","Text":"In order to find the mass of hydrogen,"},{"Start":"00:32.350 ","End":"00:34.220","Text":"we\u0027ll first find the moles of hydrogen."},{"Start":"00:34.220 ","End":"00:36.080","Text":"In order to find the moles of hydrogen,"},{"Start":"00:36.080 ","End":"00:37.970","Text":"we\u0027ll first find the moles of aluminum."},{"Start":"00:37.970 ","End":"00:40.510","Text":"So in order to find the moles of aluminum,"},{"Start":"00:40.510 ","End":"00:42.655","Text":"we\u0027ll first find the mass of aluminum."},{"Start":"00:42.655 ","End":"00:46.025","Text":"For this, we will use the equation d, density,"},{"Start":"00:46.025 ","End":"00:49.625","Text":"equals m mass divided by v, the volume."},{"Start":"00:49.625 ","End":"00:56.100","Text":"The density of aluminum is given and equals 2.70 grams per centimeter^3."},{"Start":"00:56.540 ","End":"01:03.440","Text":"We also have the dimensions of the aluminum foil."},{"Start":"01:03.440 ","End":"01:08.510","Text":"So we can calculate our volume of the aluminum foil using our dimensions,"},{"Start":"01:08.510 ","End":"01:14.180","Text":"which is 10.60 centimeters times 5.20"},{"Start":"01:14.180 ","End":"01:20.650","Text":"centimeters times 0.375 mm."},{"Start":"01:20.650 ","End":"01:23.720","Text":"Now, in order to get our volume in centimeters^3,"},{"Start":"01:23.720 ","End":"01:27.085","Text":"we will convert our millimeters to centimeters."},{"Start":"01:27.085 ","End":"01:29.205","Text":"We\u0027ll use the conversion factor."},{"Start":"01:29.205 ","End":"01:30.825","Text":"Per 1 centimeter,"},{"Start":"01:30.825 ","End":"01:32.505","Text":"we have 10 millimeters."},{"Start":"01:32.505 ","End":"01:34.200","Text":"The millimeters cancels out,"},{"Start":"01:34.200 ","End":"01:40.350","Text":"and this equals 10.60 centimeters times 5.20"},{"Start":"01:40.350 ","End":"01:47.940","Text":"centimeters times 0.0375 centimeters."},{"Start":"01:47.940 ","End":"01:55.855","Text":"This equals 2.067 centimeters^3."},{"Start":"01:55.855 ","End":"01:59.020","Text":"Now that we have our volume and our density,"},{"Start":"01:59.020 ","End":"02:01.690","Text":"we can calculate the mass of the aluminum foil."},{"Start":"02:01.690 ","End":"02:04.330","Text":"By multiplying both sides by the volume,"},{"Start":"02:04.330 ","End":"02:06.790","Text":"we get the equation m,"},{"Start":"02:06.790 ","End":"02:08.410","Text":"mass equals d,"},{"Start":"02:08.410 ","End":"02:10.450","Text":"density times v, volume."},{"Start":"02:10.450 ","End":"02:14.650","Text":"The density is 2.70 grams per centimeters^3 times"},{"Start":"02:14.650 ","End":"02:19.750","Text":"the volume which we calculated and is 2.067 centimeter^3."},{"Start":"02:19.750 ","End":"02:25.180","Text":"Centimeters^3 cancels out and we receive 5.58 grams."},{"Start":"02:25.180 ","End":"02:28.540","Text":"So the mass of the aluminum is 5.58 grams."},{"Start":"02:28.540 ","End":"02:31.390","Text":"The next step is to find the moles of the aluminum."},{"Start":"02:31.390 ","End":"02:33.280","Text":"We\u0027re going to use the equation,"},{"Start":"02:33.280 ","End":"02:35.665","Text":"n number of moles equals m,"},{"Start":"02:35.665 ","End":"02:37.705","Text":"mass divided by molar mass."},{"Start":"02:37.705 ","End":"02:40.340","Text":"Again, the number of moles equals the mass of aluminum,"},{"Start":"02:40.340 ","End":"02:42.455","Text":"which is 5.58 grams,"},{"Start":"02:42.455 ","End":"02:45.185","Text":"divided by the molar mass of aluminum,"},{"Start":"02:45.185 ","End":"02:48.470","Text":"which is 26.98 grams per mole."},{"Start":"02:48.470 ","End":"02:51.170","Text":"This value is taken from the periodic table of elements."},{"Start":"02:51.170 ","End":"02:56.165","Text":"The moles of aluminum equals 0.207 mole."},{"Start":"02:56.165 ","End":"02:59.360","Text":"Now we want to calculate the moles of hydrogen and they"},{"Start":"02:59.360 ","End":"03:03.170","Text":"equal the moles of aluminum times,"},{"Start":"03:03.170 ","End":"03:07.060","Text":"and if we look at our equation for every 2 moles of aluminum,"},{"Start":"03:07.060 ","End":"03:16.560","Text":"we have 3 moles of hydrogen so times 3 moles of hydrogen per 2 moles of aluminum."},{"Start":"03:16.560 ","End":"03:19.340","Text":"This equals 0.207 mole,"},{"Start":"03:19.340 ","End":"03:21.600","Text":"which is the moles of aluminum,"},{"Start":"03:23.300 ","End":"03:29.050","Text":"times 3 moles of hydrogen for every 2 moles of aluminum."},{"Start":"03:29.120 ","End":"03:38.070","Text":"The moles of aluminum cancel out and we get 0.31 moles of hydrogen."},{"Start":"03:38.480 ","End":"03:43.025","Text":"Our last step is to convert the moles of hydrogen into grams."},{"Start":"03:43.025 ","End":"03:45.710","Text":"For this purpose, we will use again the equation,"},{"Start":"03:45.710 ","End":"03:49.180","Text":"moles equals mass divided by molar mass."},{"Start":"03:49.180 ","End":"03:51.720","Text":"This time we want to calculate the mass."},{"Start":"03:51.720 ","End":"03:53.930","Text":"The mass equals n,"},{"Start":"03:53.930 ","End":"03:56.225","Text":"number of moles times molar mass,"},{"Start":"03:56.225 ","End":"03:58.520","Text":"just multiply both sides by the molar mass."},{"Start":"03:58.520 ","End":"04:00.335","Text":"The moles of hydrogen we know,"},{"Start":"04:00.335 ","End":"04:02.420","Text":"and they equals 0.31 moles."},{"Start":"04:02.420 ","End":"04:05.000","Text":"Now, we will calculate the molar mass of hydrogen."},{"Start":"04:05.000 ","End":"04:06.500","Text":"The molar mass of hydrogen,"},{"Start":"04:06.500 ","End":"04:11.910","Text":"which is H_2=2 times the molar mass of 1 hydrogen,"},{"Start":"04:11.910 ","End":"04:14.190","Text":"which equals 2 times 1,"},{"Start":"04:14.190 ","End":"04:17.200","Text":"which equals 2 grams per mole."},{"Start":"04:17.380 ","End":"04:21.740","Text":"Again, the mass of hydrogen equals the moles of hydrogen,"},{"Start":"04:21.740 ","End":"04:25.430","Text":"which is 0.31 mole times the molar mass of hydrogen,"},{"Start":"04:25.430 ","End":"04:27.290","Text":"which is 2 grams per mole,"},{"Start":"04:27.290 ","End":"04:33.435","Text":"moles cancel out and we\u0027re left with 0.62 grams of hydrogen."},{"Start":"04:33.435 ","End":"04:37.240","Text":"The mass of hydrogen comes to 0.62 grams."},{"Start":"04:37.240 ","End":"04:38.510","Text":"That is our final answer."},{"Start":"04:38.510 ","End":"04:41.100","Text":"Thank you very much for watching."}],"ID":21140},{"Watched":false,"Name":"Exercise 8","Duration":"2m 59s","ChapterTopicVideoID":23395,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:03.629","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.629 ","End":"00:07.440","Text":"How many moles of nitric oxide can be produced in the reaction of"},{"Start":"00:07.440 ","End":"00:12.790","Text":"5 moles of ammonia and 6 moles of oxygen?"},{"Start":"00:13.070 ","End":"00:20.380","Text":"4 ammonia plus 5 oxygen gives us 4 nitric oxide plus 6 water."},{"Start":"00:20.380 ","End":"00:25.781","Text":"So before we can calculate the number of moles of nitric oxide that are produced,"},{"Start":"00:25.781 ","End":"00:27.920","Text":"we have this decide who is the limiting reactant,"},{"Start":"00:27.920 ","End":"00:30.510","Text":"the ammonia or the oxygen."},{"Start":"00:30.510 ","End":"00:34.885","Text":"We\u0027re given 5 moles of ammonia and 6 moles of oxygen."},{"Start":"00:34.885 ","End":"00:37.670","Text":"The first step, we\u0027re going to take our 5 moles of"},{"Start":"00:37.670 ","End":"00:39.890","Text":"ammonia and calculate the number of moles"},{"Start":"00:39.890 ","End":"00:44.240","Text":"of oxygen which are needed to react with the 5 moles of ammonia."},{"Start":"00:44.240 ","End":"00:50.340","Text":"The number of moles of oxygen equal the number of moles of ammonia."},{"Start":"00:50.800 ","End":"00:57.620","Text":"If we look here, we can see that 4 ammonia react with 5 oxygen so the number of moles of"},{"Start":"00:57.620 ","End":"01:06.100","Text":"ammonia times 5 moles of oxygen divided by 4 moles of ammonia."},{"Start":"01:07.150 ","End":"01:12.880","Text":"The moles of ammonia are in the denominator since we want them to cancel out."},{"Start":"01:12.880 ","End":"01:14.570","Text":"As we said, we\u0027re given 5 moles of"},{"Start":"01:14.570 ","End":"01:22.170","Text":"ammonia times 5 moles"},{"Start":"01:22.170 ","End":"01:29.638","Text":"of oxygen divided by 4 moles of ammonia."},{"Start":"01:29.638 ","End":"01:35.360","Text":"This equals 6.25 moles of oxygen."},{"Start":"01:35.360 ","End":"01:39.014","Text":"Meaning if 5 moles of ammonia react,"},{"Start":"01:39.014 ","End":"01:43.200","Text":"we need 6.25 moles of oxygen to react with them."},{"Start":"01:43.200 ","End":"01:44.840","Text":"Now, as you can see in the question,"},{"Start":"01:44.840 ","End":"01:51.215","Text":"we only have 6 moles of oxygen so the oxygen is the limiting reactant."},{"Start":"01:51.215 ","End":"01:53.870","Text":"Now, since we know oxygen is the limiting reactant,"},{"Start":"01:53.870 ","End":"01:57.650","Text":"we\u0027re now going to calculate the number of moles of nitric oxide."},{"Start":"01:57.650 ","End":"02:01.970","Text":"The number of moles of nitric oxide equals the number of moles of"},{"Start":"02:01.970 ","End":"02:06.710","Text":"oxygen times 4 moles of nitric oxide."},{"Start":"02:06.710 ","End":"02:09.469","Text":"For every 5 moles of oxygen,"},{"Start":"02:09.469 ","End":"02:19.347","Text":"times 4 moles of nitric oxide divided by 5 moles of oxygen."},{"Start":"02:19.347 ","End":"02:22.775","Text":"As we said, the number of moles of oxygen are 6 moles."},{"Start":"02:22.775 ","End":"02:24.260","Text":"This is 6 moles of"},{"Start":"02:24.260 ","End":"02:34.121","Text":"oxygen times 4 moles of nitric oxide divided by 5 moles of oxygen."},{"Start":"02:34.121 ","End":"02:35.570","Text":"The moles of oxygen cancel"},{"Start":"02:35.570 ","End":"02:42.470","Text":"out and this equals"},{"Start":"02:42.470 ","End":"02:48.120","Text":"4.8 moles of nitric oxide."},{"Start":"02:48.190 ","End":"02:54.875","Text":"Again, the number of moles of nitric oxide equals 4.8 mole nitric oxide."},{"Start":"02:54.875 ","End":"02:56.885","Text":"That is our final answer."},{"Start":"02:56.885 ","End":"02:59.610","Text":"Thank you very much for watching."}],"ID":24284},{"Watched":false,"Name":"Exercise 9","Duration":"5m 19s","ChapterTopicVideoID":23396,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:03.540","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.540 ","End":"00:13.050","Text":"A 0.757 mole sample of copper is added to 150 millimeters of 5 molar nitric acid."},{"Start":"00:13.050 ","End":"00:16.890","Text":"Assuming the following reaction is the only one that occurs,"},{"Start":"00:16.890 ","End":"00:19.485","Text":"will the copper react completely?"},{"Start":"00:19.485 ","End":"00:29.030","Text":"3 copper plus 8 nitric acid gives us 3 copper nitrate plus 4 water plus 2 nitric oxide."},{"Start":"00:29.030 ","End":"00:33.920","Text":"We know that the number of moles of the copper which are given are 0.757 moles."},{"Start":"00:33.920 ","End":"00:43.740","Text":"That\u0027s the number of moles which are given in our question."},{"Start":"00:43.740 ","End":"00:48.095","Text":"Now we\u0027re going to calculate the number of moles which are given of the nitric acid."},{"Start":"00:48.095 ","End":"00:51.530","Text":"We know that the volume of nitric acid is"},{"Start":"00:51.530 ","End":"00:53.820","Text":"150 millimeters"},{"Start":"01:00.980 ","End":"01:02.210","Text":"and we"},{"Start":"01:02.210 ","End":"01:10.865","Text":"know that the molarity of the nitric acid equals 5 molar so that molarity equals 5 molar."},{"Start":"01:10.865 ","End":"01:14.000","Text":"We want to calculate the number of moles."},{"Start":"01:14.000 ","End":"01:18.545","Text":"The molarity equals the number of moles of the solute divided by"},{"Start":"01:18.545 ","End":"01:23.660","Text":"the volume of the solution and we want to calculate the number of moles."},{"Start":"01:23.660 ","End":"01:27.050","Text":"We\u0027re going to multiply both sides by the volume so the number of moles of"},{"Start":"01:27.050 ","End":"01:33.680","Text":"nitric acid equals the molarity times the volume."},{"Start":"01:33.680 ","End":"01:42.480","Text":"The molarity equals 5 molar and the volume is 150 millimeters."},{"Start":"01:43.130 ","End":"01:47.810","Text":"But the volume needs to be in liters so we\u0027re going to multiply by a conversion factor."},{"Start":"01:47.810 ","End":"01:52.945","Text":"We\u0027re going to multiply it by 1 liter per 1,000 millimeters."},{"Start":"01:52.945 ","End":"01:55.410","Text":"The milliliters cancel out,"},{"Start":"01:55.410 ","End":"02:01.915","Text":"and we get 0.75 moles of nitric acid."},{"Start":"02:01.915 ","End":"02:04.870","Text":"In the question, we\u0027re given the number of moles of the copper."},{"Start":"02:04.870 ","End":"02:07.775","Text":"Again, they\u0027re equal 0.757 moles."},{"Start":"02:07.775 ","End":"02:12.500","Text":"We calculated the number of moles of the nitric acid and it equals 0.75 moles."},{"Start":"02:12.500 ","End":"02:15.320","Text":"Now we have to decide who is the limiting reactant."},{"Start":"02:15.320 ","End":"02:21.060","Text":"Then we can see if we have enough copper or not for the reaction to occur."},{"Start":"02:21.880 ","End":"02:24.830","Text":"What we\u0027re going to do is we\u0027re going to take the number of moles of"},{"Start":"02:24.830 ","End":"02:27.560","Text":"the nitric acid which we calculated and we\u0027re going to"},{"Start":"02:27.560 ","End":"02:30.215","Text":"calculate the number of moles of the copper which are needed"},{"Start":"02:30.215 ","End":"02:33.160","Text":"to react with this amount of nitric acid."},{"Start":"02:33.160 ","End":"02:36.500","Text":"The number of moles of copper which are needed to react with"},{"Start":"02:36.500 ","End":"02:40.970","Text":"the number of moles of nitric acid which are given."},{"Start":"02:40.970 ","End":"02:45.770","Text":"We calculate this by multiplying this by 3 moles of"},{"Start":"02:45.770 ","End":"02:51.480","Text":"copper for every 8 moles of nitric acid."},{"Start":"02:53.080 ","End":"02:55.640","Text":"Because if we look at our reaction,"},{"Start":"02:55.640 ","End":"02:58.430","Text":"we can see for every 3 moles of copper,"},{"Start":"02:58.430 ","End":"03:01.870","Text":"8 moles of nitric acid react."},{"Start":"03:01.870 ","End":"03:04.969","Text":"The moles of nitric acid are in the denominator"},{"Start":"03:04.969 ","End":"03:08.330","Text":"since they will cancel out with moles of nitric acid."},{"Start":"03:08.330 ","End":"03:17.975","Text":"Again, the moles of nitric acid which we calculated equals 0.75 mole of"},{"Start":"03:17.975 ","End":"03:20.780","Text":"HNO_3 times 3 moles of"},{"Start":"03:20.780 ","End":"03:28.220","Text":"copper divided by 8 moles of nitric acid."},{"Start":"03:28.220 ","End":"03:34.280","Text":"The moles of nitric acid cancel out and this equals"},{"Start":"03:34.280 ","End":"03:45.815","Text":"0.28 moles of copper."},{"Start":"03:45.815 ","End":"03:50.800","Text":"If we look again, the number of moles of copper which are given is 0.757 mole."},{"Start":"03:50.800 ","End":"03:53.960","Text":"The number of moles of copper, which we calculated,"},{"Start":"03:53.960 ","End":"03:59.875","Text":"which needs to react with the nitric acid which we have are 0.28 moles of copper."},{"Start":"03:59.875 ","End":"04:06.110","Text":"Meaning that we have a lot more copper than what we actually need."},{"Start":"04:06.110 ","End":"04:10.224","Text":"In the question, will the copper react completely?"},{"Start":"04:10.224 ","End":"04:20.750","Text":"The answer is no since 0.28 moles will react and we will be left with an excess."},{"Start":"04:20.750 ","End":"04:24.510","Text":"The answer is the copper will not react completely."},{"Start":"04:42.100 ","End":"04:45.169","Text":"That\u0027s our answer. Also,"},{"Start":"04:45.169 ","End":"04:48.350","Text":"since we see that we have less moles of copper which are needed to"},{"Start":"04:48.350 ","End":"04:52.770","Text":"react with our nitric acid,"},{"Start":"04:53.260 ","End":"04:56.090","Text":"then moles of copper that we actually have,"},{"Start":"04:56.090 ","End":"05:01.500","Text":"the limiting reactant is the nitric acid, just to note."},{"Start":"05:01.640 ","End":"05:05.780","Text":"Our final answer is the copper will not react completely."},{"Start":"05:05.780 ","End":"05:08.485","Text":"Since again, we have an excess of copper,"},{"Start":"05:08.485 ","End":"05:12.170","Text":"since in the beginning we had 0.757 moles and we only"},{"Start":"05:12.170 ","End":"05:15.740","Text":"need 0.28 moles of copper to react with our nitric acid."},{"Start":"05:15.740 ","End":"05:17.165","Text":"That is our final answer."},{"Start":"05:17.165 ","End":"05:19.650","Text":"Thank you very much for watching."}],"ID":24285},{"Watched":false,"Name":"Exercise 10","Duration":"8m 39s","ChapterTopicVideoID":23394,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.729","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.729 ","End":"00:06.930","Text":"How many grams of sodium trithiocarbonate are produced in the reaction of"},{"Start":"00:06.930 ","End":"00:13.140","Text":"95 milliliters of liquid carbon disulfide density equals 1.26 grams per milliliter,"},{"Start":"00:13.140 ","End":"00:16.740","Text":"and 2.65 mole of sodium hydroxide?"},{"Start":"00:16.740 ","End":"00:20.470","Text":"3 carbon disulfide plus 6 sodium hydroxide gives us"},{"Start":"00:20.470 ","End":"00:24.845","Text":"2 sodium trithiocarbonate plus sodium carbonate plus 3 water."},{"Start":"00:24.845 ","End":"00:27.705","Text":"In this question, we\u0027re asked to find the mass,"},{"Start":"00:27.705 ","End":"00:31.105","Text":"the number of grams of the sodium thiocarbonate."},{"Start":"00:31.105 ","End":"00:34.265","Text":"For this purpose, we will first determine"},{"Start":"00:34.265 ","End":"00:38.690","Text":"the limiting reactant if it is the sodium hydroxide or the carbon disulfide."},{"Start":"00:38.690 ","End":"00:41.720","Text":"After we will decide which one is the limiting reactant,"},{"Start":"00:41.720 ","End":"00:45.695","Text":"we will calculate the moles of the sodium trithiocarbonate,"},{"Start":"00:45.695 ","End":"00:49.365","Text":"and then we will calculate the mass of the sodium trithiocarbonate."},{"Start":"00:49.365 ","End":"00:54.020","Text":"From the question we know that the sodium hydroxide has 2.65 moles."},{"Start":"00:54.020 ","End":"01:00.755","Text":"The moles of sodium hydroxide equal 2.65 mole."},{"Start":"01:00.755 ","End":"01:03.260","Text":"That\u0027s one thing we are given."},{"Start":"01:03.260 ","End":"01:06.560","Text":"The other one is the volume of the carbon disulfide so"},{"Start":"01:06.560 ","End":"01:11.145","Text":"the volume of the carbon disulfide equals 95 milliliters."},{"Start":"01:11.145 ","End":"01:14.470","Text":"We\u0027re also given the density of the carbon disulfide,"},{"Start":"01:14.470 ","End":"01:16.690","Text":"which is 1.26 grams per milliliter."},{"Start":"01:16.690 ","End":"01:22.990","Text":"The density of carbon disulfide equals 1.26 grams per milliliter."},{"Start":"01:22.990 ","End":"01:27.300","Text":"Here we\u0027ll use the equation d density equals m,"},{"Start":"01:27.300 ","End":"01:29.340","Text":"mass divided by v, the volume."},{"Start":"01:29.340 ","End":"01:33.174","Text":"Using this equation we can find the mass of the carbon disulfide."},{"Start":"01:33.174 ","End":"01:35.595","Text":"We actually multiply both sides by the volume."},{"Start":"01:35.595 ","End":"01:38.710","Text":"The mass of carbon disulfide equals"},{"Start":"01:38.710 ","End":"01:42.520","Text":"the density of carbon disulfide times the volume of carbon disulfide,"},{"Start":"01:42.520 ","End":"01:44.500","Text":"and both of these are given,"},{"Start":"01:44.500 ","End":"01:50.095","Text":"so it\u0027s 1.26 grams per milliliter times 95 milliliter."},{"Start":"01:50.095 ","End":"01:52.365","Text":"Milliliters, of course, cancel out,"},{"Start":"01:52.365 ","End":"01:55.830","Text":"and we get 119.7 grams."},{"Start":"01:55.830 ","End":"01:59.680","Text":"The next step is to find the number of moles of the carbon disulfide."},{"Start":"01:59.680 ","End":"02:03.920","Text":"That way we can compare them to the moles of the sodium hydroxide later."},{"Start":"02:03.920 ","End":"02:10.063","Text":"In order to find the number of moles we will use the equation n moles equals mass,"},{"Start":"02:10.063 ","End":"02:12.080","Text":"m, divided by molar mass."},{"Start":"02:12.080 ","End":"02:15.360","Text":"Now we need to calculate the molar mass of the carbon disulfide."},{"Start":"02:15.360 ","End":"02:20.270","Text":"The molar mass of carbon disulfide equals the molar mass of carbon since we have"},{"Start":"02:20.270 ","End":"02:27.990","Text":"1 carbon plus 2 times the molar mass of sulfur since we have 2 sulfur."},{"Start":"02:27.990 ","End":"02:37.530","Text":"This equals 12 grams per mole plus 2 times 32.1 grams per mole."},{"Start":"02:37.600 ","End":"02:41.765","Text":"The molar masses are taken from the periodic table of elements."},{"Start":"02:41.765 ","End":"02:47.190","Text":"This equals 76.2 grams per mole."},{"Start":"02:47.260 ","End":"02:52.400","Text":"Again, the moles of carbon disulfide equal"},{"Start":"02:52.400 ","End":"02:57.364","Text":"the mass of carbon disulfide divided by the molar mass of carbon disulfide."},{"Start":"02:57.364 ","End":"03:00.575","Text":"The mass is calculated and then equals 119.7"},{"Start":"03:00.575 ","End":"03:05.790","Text":"grams divided by the molar mass which is 76.2 grams per mole,"},{"Start":"03:05.790 ","End":"03:08.159","Text":"which we also calculated."},{"Start":"03:08.159 ","End":"03:13.785","Text":"This equals 1.57 moles of carbon disulfide."},{"Start":"03:13.785 ","End":"03:16.970","Text":"Now, I just want to remind you here that when you divide by"},{"Start":"03:16.970 ","End":"03:22.490","Text":"a fraction it is the same by multiplying by the reciprocal of the fraction."},{"Start":"03:22.490 ","End":"03:28.070","Text":"Grams divided by grams per mole is same as grams times mole per gram."},{"Start":"03:28.070 ","End":"03:33.570","Text":"The grams cancel out, and our units are mole, just like here."},{"Start":"03:33.570 ","End":"03:40.600","Text":"We have on one hand the moles of the sodium hydroxide which were given 2.65 mole."},{"Start":"03:41.030 ","End":"03:45.200","Text":"On the other hand, we\u0027d calculated that the moles of the carbon disulfide are"},{"Start":"03:45.200 ","End":"03:50.365","Text":"1.57 mole so let\u0027s call this 1 and 2."},{"Start":"03:50.365 ","End":"03:53.405","Text":"In order to determine which is the limiting reactant"},{"Start":"03:53.405 ","End":"03:56.780","Text":"between the sodium hydroxide and the carbon disulfide,"},{"Start":"03:56.780 ","End":"03:59.660","Text":"we\u0027ll take the 1.57 moles and we\u0027ll"},{"Start":"03:59.660 ","End":"04:03.035","Text":"see how many moles of sodium hydroxide we get for this."},{"Start":"04:03.035 ","End":"04:10.980","Text":"The moles of sodium hydroxide equal the moles of the carbon disulfide times,"},{"Start":"04:10.980 ","End":"04:12.305","Text":"if we look at our equation,"},{"Start":"04:12.305 ","End":"04:15.170","Text":"we can see that for every 3 moles of carbon disulfide which"},{"Start":"04:15.170 ","End":"04:18.215","Text":"react 6 moles of sodium hydroxide react."},{"Start":"04:18.215 ","End":"04:20.330","Text":"We need to multiply this by 6 moles of"},{"Start":"04:20.330 ","End":"04:29.030","Text":"sodium hydroxide divided by 3 moles of carbon disulfide."},{"Start":"04:29.030 ","End":"04:31.055","Text":"6 divide by 3 is of course 2."},{"Start":"04:31.055 ","End":"04:33.970","Text":"We can cancel these out and just write 2 divide by 1,"},{"Start":"04:33.970 ","End":"04:39.600","Text":"and this equals 1.57 mole of"},{"Start":"04:39.600 ","End":"04:50.155","Text":"carbon disulfide times 2 moles of sodium hydroxide divided by 1 mole of carbon disulfide."},{"Start":"04:50.155 ","End":"04:52.770","Text":"Just separate here."},{"Start":"04:52.770 ","End":"04:56.840","Text":"The moles of carbon disulfide cancel each other out and"},{"Start":"04:56.840 ","End":"05:00.575","Text":"we get 3.14 moles of sodium hydroxide."},{"Start":"05:00.575 ","End":"05:03.170","Text":"Now, if we compare in our first case,"},{"Start":"05:03.170 ","End":"05:06.785","Text":"we have moles of sodium hydroxide 2.65 moles."},{"Start":"05:06.785 ","End":"05:08.615","Text":"In our second case,"},{"Start":"05:08.615 ","End":"05:12.350","Text":"we get 3.14 moles of sodium hydroxide."},{"Start":"05:12.350 ","End":"05:16.385","Text":"Our first case is the limiting reactant because we don\u0027t have"},{"Start":"05:16.385 ","End":"05:23.585","Text":"enough sodium hydroxide to react with the 1.57 moles of carbon disulfide."},{"Start":"05:23.585 ","End":"05:29.930","Text":"Therefore, the moles of sodium hydroxide are 2.65 mole and it is the limiting reactant."},{"Start":"05:29.930 ","End":"05:33.260","Text":"Now that we know that the sodium hydroxide is the limiting reactant,"},{"Start":"05:33.260 ","End":"05:37.670","Text":"we have to find the moles of the sodium trithiocarbonate."},{"Start":"05:37.670 ","End":"05:45.875","Text":"The moles of the sodium trithiocarbonate equal the moles of the sodium hydroxide times,"},{"Start":"05:45.875 ","End":"05:50.300","Text":"if we look at our equation we can see that for every 6 moles of sodium hydroxide which"},{"Start":"05:50.300 ","End":"05:54.825","Text":"react we get 2 moles of sodium trithiocarbonate."},{"Start":"05:54.825 ","End":"05:56.880","Text":"We can multiply this by 2 moles of"},{"Start":"05:56.880 ","End":"06:03.010","Text":"sodium trithiocarbonate divided by 6 moles of sodium hydroxide."},{"Start":"06:06.290 ","End":"06:13.870","Text":"The 2 and the 6, we can divide them by 2 and we get 1-3 instead of 2-6."},{"Start":"06:13.870 ","End":"06:17.690","Text":"The moles of the sodium hydroxide will cancel out."},{"Start":"06:17.690 ","End":"06:20.705","Text":"We get the moles of the sodium hydroxide,"},{"Start":"06:20.705 ","End":"06:28.215","Text":"which we said is 2.65 mole of sodium hydroxide times"},{"Start":"06:28.215 ","End":"06:34.005","Text":"1 mole of the sodium trithiocarbonate"},{"Start":"06:34.005 ","End":"06:40.905","Text":"divided by 3 moles of the sodium hydroxide."},{"Start":"06:40.905 ","End":"06:42.890","Text":"We\u0027ll cancel the moles of sodium hydroxide right"},{"Start":"06:42.890 ","End":"06:45.440","Text":"now because it\u0027s more comfortable to do it this way."},{"Start":"06:45.440 ","End":"06:50.500","Text":"Here we are left with 0.88 moles sodium trithiocarbonate."},{"Start":"06:50.500 ","End":"06:53.240","Text":"In this step,"},{"Start":"06:53.240 ","End":"06:55.700","Text":"we found the moles of the sodium trithiocarbonate to"},{"Start":"06:55.700 ","End":"06:58.400","Text":"the moles of sodium trithiocarbonate."},{"Start":"06:58.400 ","End":"07:03.320","Text":"Now the last step is to calculate the mass of sodium trithiocarbonate."},{"Start":"07:03.320 ","End":"07:06.125","Text":"Recall that n mole,"},{"Start":"07:06.125 ","End":"07:09.370","Text":"equals m mass divided by molar mass."},{"Start":"07:09.370 ","End":"07:11.780","Text":"In order to find the mass first, we need the molar mass."},{"Start":"07:11.780 ","End":"07:17.705","Text":"The molar mass of the sodium trithiocarbonate equals 2 times the molar mass"},{"Start":"07:17.705 ","End":"07:24.335","Text":"of the sodium plus the molar mass of carbon plus 3 times the molar mass of sulfur,"},{"Start":"07:24.335 ","End":"07:31.235","Text":"and this equals 2 times 23 grams per mole plus"},{"Start":"07:31.235 ","End":"07:39.900","Text":"12 grams per mole plus 3 times 32.1 grams per mole."},{"Start":"07:40.960 ","End":"07:47.080","Text":"This comes to 154.3 grams per mole."},{"Start":"07:47.080 ","End":"07:51.290","Text":"Of course, all the molar masses were taken from the periodic table of elements."},{"Start":"07:51.290 ","End":"07:53.180","Text":"Now we can find the mass."},{"Start":"07:53.180 ","End":"07:55.790","Text":"If we multiply both sides by the molar mass,"},{"Start":"07:55.790 ","End":"07:58.440","Text":"we can see that the mass of"},{"Start":"07:59.150 ","End":"08:10.090","Text":"sodium trithiocarbonate equals the moles times the molar mass."},{"Start":"08:13.790 ","End":"08:22.010","Text":"Now the moles are 0.88 mole times the molar mass,"},{"Start":"08:22.010 ","End":"08:26.670","Text":"which is 154.3 grams per mole."},{"Start":"08:26.770 ","End":"08:30.650","Text":"Here the moles cancel out and our final answer comes to"},{"Start":"08:30.650 ","End":"08:35.810","Text":"135.78 grams of sodium trithiocarbonate."},{"Start":"08:35.810 ","End":"08:37.415","Text":"That is our final answer."},{"Start":"08:37.415 ","End":"08:40.110","Text":"Thank you so much for watching."}],"ID":24283},{"Watched":false,"Name":"Exercise 11","Duration":"5m 54s","ChapterTopicVideoID":20332,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.210 ","End":"00:06.975","Text":"What are the molarities of the following solutes when dissolved in water."},{"Start":"00:06.975 ","End":"00:14.205","Text":"A, 3.11 mole of methanol in 6.45 liters of solution, b,"},{"Start":"00:14.205 ","End":"00:20.130","Text":"8.26 millimole of ethanol in 45 milliliters of solution,"},{"Start":"00:20.130 ","End":"00:27.000","Text":"or c, 23.5 grams of urea in 311 milliliters of solution."},{"Start":"00:27.000 ","End":"00:32.610","Text":"We will begin with a. Now recall that the molarity, M,"},{"Start":"00:32.610 ","End":"00:37.020","Text":"equals the amount of solute in moles,"},{"Start":"00:37.020 ","End":"00:39.250","Text":"so moles of solute,"},{"Start":"00:39.250 ","End":"00:44.390","Text":"divided by V, which is the volume of the solution in liters."},{"Start":"00:44.390 ","End":"00:48.270","Text":"Solution in liters. In a,"},{"Start":"00:48.270 ","End":"00:52.265","Text":"the molarity equals the moles of solute,"},{"Start":"00:52.265 ","End":"00:54.095","Text":"which is 3.11 mole,"},{"Start":"00:54.095 ","End":"00:57.980","Text":"divided by the volume of the solution in liters,"},{"Start":"00:57.980 ","End":"01:00.215","Text":"which is 6.45 liters."},{"Start":"01:00.215 ","End":"01:06.540","Text":"This equals 0.482 molar."},{"Start":"01:06.540 ","End":"01:09.510","Text":"That is our final answer for a."},{"Start":"01:09.510 ","End":"01:12.990","Text":"In b, we will use the same equation."},{"Start":"01:12.990 ","End":"01:16.700","Text":"Again, molarity equals the amount of"},{"Start":"01:16.700 ","End":"01:20.450","Text":"solute in moles divided by the volume of the solution in liters."},{"Start":"01:20.450 ","End":"01:25.135","Text":"In b, the moles of solute equals 8.26 millimole."},{"Start":"01:25.135 ","End":"01:35.120","Text":"Molarity equals 8.26 millimole divided by the volume of the solution,"},{"Start":"01:35.120 ","End":"01:37.445","Text":"which is 45 milliliters."},{"Start":"01:37.445 ","End":"01:42.680","Text":"Now we want the amount of the solute to be in moles."},{"Start":"01:42.680 ","End":"01:46.025","Text":"Therefore, we will multiply by a conversion factor."},{"Start":"01:46.025 ","End":"01:47.945","Text":"For every 1 mole,"},{"Start":"01:47.945 ","End":"01:50.885","Text":"we have 1,000 millimoles."},{"Start":"01:50.885 ","End":"01:53.285","Text":"The millimoles cancel out,"},{"Start":"01:53.285 ","End":"02:01.740","Text":"and we\u0027re left with 8.26 times 10 to the negative 3 mole in our numerator."},{"Start":"02:01.740 ","End":"02:05.405","Text":"In the denominator we have 45 milliliters."},{"Start":"02:05.405 ","End":"02:07.955","Text":"We need the volume of the solution to be in liters."},{"Start":"02:07.955 ","End":"02:11.005","Text":"So we will multiply by a conversion factor."},{"Start":"02:11.005 ","End":"02:13.170","Text":"For every 1 liter,"},{"Start":"02:13.170 ","End":"02:15.265","Text":"we have 1,000 milliliters."},{"Start":"02:15.265 ","End":"02:17.270","Text":"The milliliters cancel out,"},{"Start":"02:17.270 ","End":"02:22.865","Text":"and in our denominator we have 0.045 liters."},{"Start":"02:22.865 ","End":"02:31.275","Text":"Now, this equals 0.184 mole per liter."},{"Start":"02:31.275 ","End":"02:35.345","Text":"Mole per liter equals molarity or molar."},{"Start":"02:35.345 ","End":"02:40.110","Text":"This also equals 0.184 molar."},{"Start":"02:40.110 ","End":"02:45.435","Text":"Our final answer for b is 0.184 molar."},{"Start":"02:45.435 ","End":"02:50.700","Text":"Now we will continue to c. In c we have 23.5"},{"Start":"02:50.700 ","End":"02:56.640","Text":"grams of solute in 311 milliliters of solution."},{"Start":"02:56.640 ","End":"02:59.080","Text":"In c, we will also use the equation,"},{"Start":"02:59.080 ","End":"03:01.700","Text":"molarity equals amount of solute in"},{"Start":"03:01.700 ","End":"03:04.850","Text":"moles divided by the volume of the solution in liters."},{"Start":"03:04.850 ","End":"03:08.900","Text":"We will also use another equation."},{"Start":"03:08.900 ","End":"03:16.275","Text":"Recall that the number of moles equals the mass divided by the molar mass."},{"Start":"03:16.275 ","End":"03:19.670","Text":"In c, we are given the mass in grams."},{"Start":"03:19.670 ","End":"03:21.680","Text":"We can calculate the molar mass,"},{"Start":"03:21.680 ","End":"03:24.260","Text":"and then we will calculate the moles."},{"Start":"03:24.260 ","End":"03:30.680","Text":"The molar mass of urea equals the molar mass of carbon,"},{"Start":"03:30.680 ","End":"03:35.000","Text":"because we have 1 carbon plus the molar mass of oxygen,"},{"Start":"03:35.000 ","End":"03:40.564","Text":"since we have 1 oxygen plus 2 times the molar mass of nitrogen,"},{"Start":"03:40.564 ","End":"03:46.085","Text":"because we have 2 nitrogens plus 4 times the molar mass of hydrogen,"},{"Start":"03:46.085 ","End":"03:49.075","Text":"since we have 4 hydrogens here, 2 times 2."},{"Start":"03:49.075 ","End":"03:51.935","Text":"This equals the molar mass of carbon,"},{"Start":"03:51.935 ","End":"03:54.560","Text":"which is 12 plus 16,"},{"Start":"03:54.560 ","End":"03:56.105","Text":"which is the molar mass of oxygen,"},{"Start":"03:56.105 ","End":"03:59.820","Text":"plus 2 times 14,"},{"Start":"03:59.820 ","End":"04:01.080","Text":"the molar mass of nitrogen,"},{"Start":"04:01.080 ","End":"04:02.700","Text":"plus 4 times 1,"},{"Start":"04:02.700 ","End":"04:04.515","Text":"the molar mass of hydrogen."},{"Start":"04:04.515 ","End":"04:07.970","Text":"Now, all these masses can be found in the periodic table of elements."},{"Start":"04:07.970 ","End":"04:11.375","Text":"This equals 60 grams per mole."},{"Start":"04:11.375 ","End":"04:14.720","Text":"The molar mass of urea is 60 grams per mole."},{"Start":"04:14.720 ","End":"04:18.505","Text":"Now we can calculate the moles of urea."},{"Start":"04:18.505 ","End":"04:21.275","Text":"N the moles equals the mass,"},{"Start":"04:21.275 ","End":"04:26.690","Text":"which is given 23.5 grams divided by the molar mass of urea,"},{"Start":"04:26.690 ","End":"04:28.745","Text":"which is 60 grams per mole."},{"Start":"04:28.745 ","End":"04:35.845","Text":"This equals 0.392 mole."},{"Start":"04:35.845 ","End":"04:38.725","Text":"Now I just want to give a quick reminder here."},{"Start":"04:38.725 ","End":"04:40.990","Text":"When we divide by a fraction,"},{"Start":"04:40.990 ","End":"04:43.760","Text":"for example, grams divided by grams per mole,"},{"Start":"04:43.760 ","End":"04:48.015","Text":"this is the same as multiplying by the reciprocal of the fraction,"},{"Start":"04:48.015 ","End":"04:50.445","Text":"multiplying by mole divided by grams."},{"Start":"04:50.445 ","End":"04:53.160","Text":"Grams cancels out and we\u0027re left with mole."},{"Start":"04:53.160 ","End":"04:56.615","Text":"The moles of urea are 0.392 moles."},{"Start":"04:56.615 ","End":"04:59.630","Text":"Now we can use the other equation, the Molarity equation."},{"Start":"04:59.630 ","End":"05:04.465","Text":"The molarity of urea equals the moles of the solute,"},{"Start":"05:04.465 ","End":"05:06.090","Text":"which we just calculated,"},{"Start":"05:06.090 ","End":"05:08.870","Text":"divided by the volume of the solution."},{"Start":"05:08.870 ","End":"05:15.680","Text":"The moles equal 0.392 moles divided by the volume of the solution,"},{"Start":"05:15.680 ","End":"05:17.680","Text":"which is 311 milliliters."},{"Start":"05:17.680 ","End":"05:20.585","Text":"Now, since the volume of the solution should be in liters,"},{"Start":"05:20.585 ","End":"05:22.655","Text":"we will multiply by a conversion factor."},{"Start":"05:22.655 ","End":"05:24.140","Text":"In every 1 liter,"},{"Start":"05:24.140 ","End":"05:26.090","Text":"there are 1,000 milliliters."},{"Start":"05:26.090 ","End":"05:30.510","Text":"The milliliters cancel out and we\u0027re left with 0.392 in"},{"Start":"05:30.510 ","End":"05:37.730","Text":"the numerator moles divided by 0.311 liters in the denominator."},{"Start":"05:37.730 ","End":"05:42.630","Text":"This equals 1.26 molar."},{"Start":"05:42.630 ","End":"05:44.330","Text":"I reminded you here,"},{"Start":"05:44.330 ","End":"05:45.710","Text":"in case you don\u0027t remember,"},{"Start":"05:45.710 ","End":"05:49.460","Text":"that mole per liter equals molar."},{"Start":"05:49.460 ","End":"05:54.630","Text":"That is our final answer for c. Thank you very much for watching."}],"ID":21143},{"Watched":false,"Name":"Exercise 12","Duration":"8m 11s","ChapterTopicVideoID":20333,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.924","Text":"Hi. We\u0027re going to answer the following exercise."},{"Start":"00:02.924 ","End":"00:05.130","Text":"What is the molarity of, a,"},{"Start":"00:05.130 ","End":"00:10.365","Text":"sucrose C_12H_22O_11 if 250 grams"},{"Start":"00:10.365 ","End":"00:14.610","Text":"of sucrose is dissolved per 350 milliliters of water solution?"},{"Start":"00:14.610 ","End":"00:18.150","Text":"B, urea, CO(NH_2)_2,"},{"Start":"00:18.150 ","End":"00:20.850","Text":"if 95.5 milligrams of"},{"Start":"00:20.850 ","End":"00:26.430","Text":"the 97 percent pure solid is dissolved in 4 milliliters of aqueous solution?"},{"Start":"00:26.430 ","End":"00:28.350","Text":"We will begin with a."},{"Start":"00:28.350 ","End":"00:31.530","Text":"In a, we want to find the molarity of sucrose."},{"Start":"00:31.530 ","End":"00:35.100","Text":"Recall that molarity equals n,"},{"Start":"00:35.100 ","End":"00:37.560","Text":"which is the amount of solute in moles."},{"Start":"00:37.560 ","End":"00:43.575","Text":"Solute in moles divided by the volume of the solution in liters."},{"Start":"00:43.575 ","End":"00:45.695","Text":"In order to find the molarity of sucrose,"},{"Start":"00:45.695 ","End":"00:48.140","Text":"we first need to find the moles of sucrose."},{"Start":"00:48.140 ","End":"00:49.475","Text":"In order to do this,"},{"Start":"00:49.475 ","End":"00:51.800","Text":"we will use the mass of sucrose."},{"Start":"00:51.800 ","End":"00:54.895","Text":"Another equation that we need is n,"},{"Start":"00:54.895 ","End":"01:00.905","Text":"the moles equals mass m divided by the molar mass."},{"Start":"01:00.905 ","End":"01:04.595","Text":"In order to calculate the moles of sucrose,"},{"Start":"01:04.595 ","End":"01:07.680","Text":"first of all, we have the mass which is 250 grams."},{"Start":"01:07.680 ","End":"01:09.605","Text":"We have to calculate the molar mass."},{"Start":"01:09.605 ","End":"01:17.195","Text":"The molar mass of sucrose equals the molar mass of carbon times 12."},{"Start":"01:17.195 ","End":"01:24.855","Text":"Since we have 12 carbons plus 22 times the molar mass of hydrogen,"},{"Start":"01:24.855 ","End":"01:30.785","Text":"since we have 22 hydrogens plus 11 times the molar mass of oxygen,"},{"Start":"01:30.785 ","End":"01:32.890","Text":"since we have 11 oxygens."},{"Start":"01:32.890 ","End":"01:39.210","Text":"This equals 12 times 12 plus 22 times 1 plus 11 times 16."},{"Start":"01:39.210 ","End":"01:41.010","Text":"The molar mass is 12, 1,"},{"Start":"01:41.010 ","End":"01:43.245","Text":"and 16 are found in the periodic table."},{"Start":"01:43.245 ","End":"01:46.840","Text":"This equals 342 grams per mole."},{"Start":"01:46.840 ","End":"01:50.920","Text":"The molar mass of sucrose is 342 grams per mole."},{"Start":"01:50.920 ","End":"01:53.510","Text":"Now, we can calculate the moles of sucrose."},{"Start":"01:53.510 ","End":"01:56.285","Text":"The moles of sucrose equals the mass of sucrose,"},{"Start":"01:56.285 ","End":"02:00.575","Text":"which is 250 grams divided by the molar mass of sucrose,"},{"Start":"02:00.575 ","End":"02:01.955","Text":"which we calculated,"},{"Start":"02:01.955 ","End":"02:04.875","Text":"and is 342 grams per mole."},{"Start":"02:04.875 ","End":"02:09.115","Text":"This equals 0.731 mole."},{"Start":"02:09.115 ","End":"02:12.725","Text":"Now, I just want to remind you here that when you divide by a fraction,"},{"Start":"02:12.725 ","End":"02:15.265","Text":"for example, grams divided by grams per mole,"},{"Start":"02:15.265 ","End":"02:18.770","Text":"this equals grams times the reciprocal of"},{"Start":"02:18.770 ","End":"02:21.590","Text":"the fraction moles per grams because"},{"Start":"02:21.590 ","End":"02:24.725","Text":"it is the same as multiplying by the reciprocal of the fraction."},{"Start":"02:24.725 ","End":"02:28.085","Text":"Grams cancel out in our case and this equals moles."},{"Start":"02:28.085 ","End":"02:31.630","Text":"The moles of sucrose are 0.731 mole."},{"Start":"02:31.630 ","End":"02:34.350","Text":"Now, we can use the molarity equation."},{"Start":"02:34.350 ","End":"02:39.830","Text":"The molarity equals moles of solute divided by liters of solution,"},{"Start":"02:39.830 ","End":"02:41.405","Text":"which equals the moles of solute,"},{"Start":"02:41.405 ","End":"02:48.410","Text":"which is 0.731 moles divided by the volume of the solution,"},{"Start":"02:48.410 ","End":"02:52.580","Text":"which equals 350 milliliters and our volume needs to be in liters."},{"Start":"02:52.580 ","End":"02:54.440","Text":"We\u0027re going to multiply by a conversion factor."},{"Start":"02:54.440 ","End":"02:58.780","Text":"For every liter, we have 1,000 milliliters."},{"Start":"02:58.780 ","End":"03:00.968","Text":"Milliliters cancel out,"},{"Start":"03:00.968 ","End":"03:06.525","Text":"and this equals 0.731 mole divided by 0.350 liters."},{"Start":"03:06.525 ","End":"03:11.210","Text":"This equals 2.09 and mole per liter equals molar."},{"Start":"03:11.210 ","End":"03:16.175","Text":"Our final answer for a is 2.09 molar."},{"Start":"03:16.175 ","End":"03:20.155","Text":"Thank you very much for watching and we will answer b."},{"Start":"03:20.155 ","End":"03:23.420","Text":"We solved a, now we will solve b."},{"Start":"03:23.420 ","End":"03:25.865","Text":"What is the molarity of urea,"},{"Start":"03:25.865 ","End":"03:33.040","Text":"if 95.5 milligrams of the 97 percent pure solid is dissolved in 4 ml of aqueous solution?"},{"Start":"03:33.040 ","End":"03:34.610","Text":"In order to find the molarity,"},{"Start":"03:34.610 ","End":"03:39.275","Text":"we use the equation molarity M equals n,"},{"Start":"03:39.275 ","End":"03:40.790","Text":"which is the moles of solute,"},{"Start":"03:40.790 ","End":"03:43.880","Text":"so solute and mole divided by V,"},{"Start":"03:43.880 ","End":"03:47.630","Text":"which is the volume of the solution in liters."},{"Start":"03:47.630 ","End":"03:49.504","Text":"Now from the question,"},{"Start":"03:49.504 ","End":"03:53.975","Text":"we know that our solid is 95.5 milligrams,"},{"Start":"03:53.975 ","End":"03:58.475","Text":"and we know the 97 percent of the pure solid is urea."},{"Start":"03:58.475 ","End":"04:00.290","Text":"In order to find the moles,"},{"Start":"04:00.290 ","End":"04:01.655","Text":"we will use the equation,"},{"Start":"04:01.655 ","End":"04:04.370","Text":"n moles equals m,"},{"Start":"04:04.370 ","End":"04:06.980","Text":"which is the mass divided by the molar mass."},{"Start":"04:06.980 ","End":"04:09.680","Text":"Since we want to find the moles of urea,"},{"Start":"04:09.680 ","End":"04:12.695","Text":"we need the mass of urea and the molar mass of urea."},{"Start":"04:12.695 ","End":"04:16.350","Text":"The molar mass of CO(NH_2)_2,"},{"Start":"04:17.230 ","End":"04:22.715","Text":"equals the molar mass of carbon plus the molar mass of oxygen."},{"Start":"04:22.715 ","End":"04:28.835","Text":"Because we have 1 carbon and 1 oxygen plus 2 times the molar mass of nitrogen."},{"Start":"04:28.835 ","End":"04:30.635","Text":"Since we have 2 nitrogens,"},{"Start":"04:30.635 ","End":"04:35.135","Text":"nitrogen times 2 plus 4 times the molar mass of hydrogen."},{"Start":"04:35.135 ","End":"04:38.175","Text":"Since we have H_2 times 2 equals 4."},{"Start":"04:38.175 ","End":"04:44.070","Text":"This equals 12 plus 16 plus 2 times 14 plus 4 times 1."},{"Start":"04:44.070 ","End":"04:47.040","Text":"This comes to 60 grams per mole."},{"Start":"04:47.040 ","End":"04:52.120","Text":"The molar masses that were given here are taken from the periodic table of elements."},{"Start":"04:52.120 ","End":"04:56.500","Text":"I found the molar mass of urea and we have it, 60 grams per mole."},{"Start":"04:56.500 ","End":"04:59.480","Text":"Now, we need to find the mass of urea."},{"Start":"04:59.480 ","End":"05:01.210","Text":"As we know from the question,"},{"Start":"05:01.210 ","End":"05:05.905","Text":"the mass of the solid equals 95.5 milligrams."},{"Start":"05:05.905 ","End":"05:09.850","Text":"However, the solid is only 97 percent urea."},{"Start":"05:09.850 ","End":"05:11.980","Text":"In order to find the mass of urea,"},{"Start":"05:11.980 ","End":"05:20.170","Text":"you will take the mass of the total solid and multiply it by a conversion factor."},{"Start":"05:20.170 ","End":"05:22.920","Text":"Since for every 100 grams of solid,"},{"Start":"05:22.920 ","End":"05:26.140","Text":"we have 97 grams of urea to the milligrams of the"},{"Start":"05:26.140 ","End":"05:31.715","Text":"solid and this is our solid and here we have 97 grams of urea."},{"Start":"05:31.715 ","End":"05:34.550","Text":"Now, we will convert our milligrams to grams,"},{"Start":"05:34.550 ","End":"05:39.080","Text":"95.5 milligrams, in order to convert to grams,"},{"Start":"05:39.080 ","End":"05:41.840","Text":"we\u0027re going to multiply it by a conversion factor."},{"Start":"05:41.840 ","End":"05:45.240","Text":"For every 1 gram, we have 1,000 milligrams."},{"Start":"05:45.240 ","End":"05:49.670","Text":"The milligrams cancel out and we multiply by 97 grams of"},{"Start":"05:49.670 ","End":"05:55.650","Text":"urea divided by 100 grams of solid."},{"Start":"05:55.650 ","End":"05:59.325","Text":"This of course, is grams of solid also."},{"Start":"05:59.325 ","End":"06:06.275","Text":"This equals 0.0955 grams"},{"Start":"06:06.275 ","End":"06:13.620","Text":"of solid times 97 grams of urea divided by 100 grams of solid."},{"Start":"06:13.700 ","End":"06:23.015","Text":"Grams of solid cancels out and our mass comes to 0.0926 grams of urea."},{"Start":"06:23.015 ","End":"06:26.695","Text":"Now, that we have our mass and our molar mass,"},{"Start":"06:26.695 ","End":"06:28.405","Text":"we can calculate the mass,"},{"Start":"06:28.405 ","End":"06:31.990","Text":"which is 0.0926 grams divided by the molar mass,"},{"Start":"06:31.990 ","End":"06:38.270","Text":"which is 60 grams per mole and this equals 0.00154 mole."},{"Start":"06:38.270 ","End":"06:40.375","Text":"Now, I just want to remind you here,"},{"Start":"06:40.375 ","End":"06:42.860","Text":"we have grams divided by grams per mole,"},{"Start":"06:42.860 ","End":"06:45.019","Text":"so we\u0027re actually dividing by a fraction."},{"Start":"06:45.019 ","End":"06:50.320","Text":"Dividing by a fraction is the same as multiplying by the reciprocal of the fraction."},{"Start":"06:50.320 ","End":"06:51.820","Text":"Grams times mole per gram,"},{"Start":"06:51.820 ","End":"06:53.960","Text":"grams cancel out and we\u0027re left with mole."},{"Start":"06:53.960 ","End":"06:55.525","Text":"Now, that we have the number of moles,"},{"Start":"06:55.525 ","End":"06:58.715","Text":"we can use the equation that we mentioned earlier,"},{"Start":"06:58.715 ","End":"07:00.800","Text":"which is molarity equals n,"},{"Start":"07:00.800 ","End":"07:03.590","Text":"which is the moles of solute divided by V,"},{"Start":"07:03.590 ","End":"07:06.170","Text":"which is the volume of the solution in liters."},{"Start":"07:06.170 ","End":"07:08.465","Text":"The molarity equals n,"},{"Start":"07:08.465 ","End":"07:15.725","Text":"which is 0.00154 mole divided by the volume of the solution in liters."},{"Start":"07:15.725 ","End":"07:19.445","Text":"Now, the volume that was given in the question is 4 milliliters."},{"Start":"07:19.445 ","End":"07:21.020","Text":"Since we need this in liters,"},{"Start":"07:21.020 ","End":"07:23.270","Text":"we will multiply it by a conversion factor,"},{"Start":"07:23.270 ","End":"07:26.600","Text":"which for every 1 liter we have 1,000 milliliters."},{"Start":"07:26.600 ","End":"07:35.630","Text":"Our milliliters cancel out and this equals 0.00154 mole divided by 0.004"},{"Start":"07:35.630 ","End":"07:40.670","Text":"liters and this"},{"Start":"07:40.670 ","End":"07:47.410","Text":"equals 0.386 molar."},{"Start":"07:47.410 ","End":"07:52.680","Text":"I just want to remind you that mole divided by liters equals molar."},{"Start":"07:58.240 ","End":"08:04.110","Text":"The molarity of urea came out to 0.386 molar."},{"Start":"08:05.290 ","End":"08:07.745","Text":"That\u0027s our final answer."},{"Start":"08:07.745 ","End":"08:10.440","Text":"Thank you very much for watching."}],"ID":21144},{"Watched":false,"Name":"Exercise 13","Duration":"3m 20s","ChapterTopicVideoID":20334,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"Hi, we\u0027re going to answer the following exercise."},{"Start":"00:03.060 ","End":"00:04.830","Text":"What is the molarity of methanol;"},{"Start":"00:04.830 ","End":"00:08.708","Text":"density of methanol equals 0.792 grams per milliliter,"},{"Start":"00:08.708 ","End":"00:13.830","Text":"if 150 milliliters of methanol is dissolved per 20 liters of water solution?"},{"Start":"00:13.830 ","End":"00:15.600","Text":"To find the molarity,"},{"Start":"00:15.600 ","End":"00:17.220","Text":"we will use the equation molarity,"},{"Start":"00:17.220 ","End":"00:21.908","Text":"M equals amount of solute in moles to solid in moles,"},{"Start":"00:21.908 ","End":"00:24.750","Text":"divided by the volume of the solution in liters."},{"Start":"00:24.750 ","End":"00:26.865","Text":"The volume of the solution in liters."},{"Start":"00:26.865 ","End":"00:30.915","Text":"Now, in order to find the moles of solute we will use the equation n"},{"Start":"00:30.915 ","End":"00:35.910","Text":"moles equals m which is the mass divided by the molar mass."},{"Start":"00:35.910 ","End":"00:41.200","Text":"In order to find the moles of methanol we will first find the molar mass of methanol."},{"Start":"00:41.200 ","End":"00:46.960","Text":"The molar mass of methanol equals the molar mass of carbon because we have 1 carbon,"},{"Start":"00:46.960 ","End":"00:50.255","Text":"plus 4 times the molar mass of hydrogen since we have"},{"Start":"00:50.255 ","End":"00:55.267","Text":"3 hydrogens plus 1 and equals 4 plus the molar mass of oxygen,"},{"Start":"00:55.267 ","End":"00:56.895","Text":"again, we have 1 oxygen."},{"Start":"00:56.895 ","End":"01:00.720","Text":"So this equals 12 plus 4 times 1 plus 16."},{"Start":"01:00.720 ","End":"01:03.305","Text":"The molar masses can be found in the periodic table."},{"Start":"01:03.305 ","End":"01:06.920","Text":"This equals 32 grams per mole."},{"Start":"01:06.920 ","End":"01:10.460","Text":"The molar mass of methanol equals 32 grams per mole."},{"Start":"01:10.460 ","End":"01:14.945","Text":"Now we need to find the mass of methanol in order to calculate the moles."},{"Start":"01:14.945 ","End":"01:19.925","Text":"Here we are given the density of methanol which equals 0.792 grams per milliliter."},{"Start":"01:19.925 ","End":"01:23.240","Text":"Recall that the density equals the mass"},{"Start":"01:23.240 ","End":"01:27.200","Text":"divided by the volume and since we\u0027re looking for the mass,"},{"Start":"01:27.200 ","End":"01:29.380","Text":"we\u0027ll multiply both sides by the volume."},{"Start":"01:29.380 ","End":"01:31.280","Text":"The mass equals d,"},{"Start":"01:31.280 ","End":"01:33.530","Text":"the density times v, the volume."},{"Start":"01:33.530 ","End":"01:36.950","Text":"The mass equals d, density times v,"},{"Start":"01:36.950 ","End":"01:41.720","Text":"volume and the mass of methanol equals the density of methanol which is"},{"Start":"01:41.720 ","End":"01:49.620","Text":"0.792 grams per mil times the volume of methanol which is 150 milliliters."},{"Start":"01:52.340 ","End":"01:55.110","Text":"The milliliters cancel out,"},{"Start":"01:55.110 ","End":"02:00.885","Text":"so this equals 118.8 grams."},{"Start":"02:00.885 ","End":"02:03.715","Text":"The mass of methanol is 118.8 grams."},{"Start":"02:03.715 ","End":"02:06.185","Text":"Now I can calculate the moles of methanol."},{"Start":"02:06.185 ","End":"02:09.710","Text":"Again, the moles of methanol equals the mass of methanol divided"},{"Start":"02:09.710 ","End":"02:13.475","Text":"by the molar mass of methanol which equals"},{"Start":"02:13.475 ","End":"02:21.570","Text":"the mass which is 118.8 grams divided by the molar mass which is 32 grams per mole,"},{"Start":"02:23.390 ","End":"02:27.550","Text":"and this equals 3.71 mole."},{"Start":"02:27.550 ","End":"02:29.000","Text":"Now, just a quick reminder,"},{"Start":"02:29.000 ","End":"02:30.980","Text":"when you are dividing by a fraction,"},{"Start":"02:30.980 ","End":"02:34.655","Text":"it is the same as multiplying by the reciprocal of the fraction."},{"Start":"02:34.655 ","End":"02:42.310","Text":"Grams divided by grams per mole equals the same as grams times mole per gram."},{"Start":"02:43.400 ","End":"02:47.175","Text":"Grams cancel out and we get mole,"},{"Start":"02:47.175 ","End":"02:50.935","Text":"so the moles of methanol are 3.71 mole."},{"Start":"02:50.935 ","End":"02:53.765","Text":"Now we can calculate the molarity of methanol."},{"Start":"02:53.765 ","End":"02:57.950","Text":"The molarity, M equals the moles which is 3.71"},{"Start":"02:57.950 ","End":"03:02.370","Text":"mole divided by the volume of the solution in liters."},{"Start":"03:02.370 ","End":"03:08.720","Text":"This is 20 liters and this equals 0.186 molar."},{"Start":"03:08.720 ","End":"03:13.375","Text":"Since remember that mole divided by liters equals molar."},{"Start":"03:13.375 ","End":"03:18.020","Text":"Our final answer for this question is 0.186 mole."},{"Start":"03:18.020 ","End":"03:20.700","Text":"Thank you very much for watching."}],"ID":21145},{"Watched":false,"Name":"Exercise 14","Duration":"6m 47s","ChapterTopicVideoID":20335,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.804","Text":"Hi, we\u0027re going to answer the following exercise."},{"Start":"00:02.804 ","End":"00:07.095","Text":"How much a, glucose in grams must be dissolved in water to produce"},{"Start":"00:07.095 ","End":"00:12.375","Text":"80 milliliters of 0.250 molar glucose and b,"},{"Start":"00:12.375 ","End":"00:17.760","Text":"methanol density equals 0.79 grams per ml in milliliters must be"},{"Start":"00:17.760 ","End":"00:23.765","Text":"dissolved in water to produce 3.15 liters of 0.375 molar methanol."},{"Start":"00:23.765 ","End":"00:25.710","Text":"We\u0027ll begin with a. Now in a,"},{"Start":"00:25.710 ","End":"00:30.750","Text":"we\u0027re giving the volume of the solution of glucose and the volume is 80 milliliters."},{"Start":"00:30.750 ","End":"00:33.405","Text":"We\u0027re also given the molarity."},{"Start":"00:33.405 ","End":"00:37.710","Text":"Molarity equals 0.250 molar."},{"Start":"00:37.710 ","End":"00:39.090","Text":"In order to find the grams,"},{"Start":"00:39.090 ","End":"00:40.238","Text":"we will first find the moles,"},{"Start":"00:40.238 ","End":"00:44.060","Text":"and we will use the equation m molarity equals n,"},{"Start":"00:44.060 ","End":"00:45.560","Text":"which is the moles of solute,"},{"Start":"00:45.560 ","End":"00:49.190","Text":"divided by V, which is the volume of the solution in liters."},{"Start":"00:49.190 ","End":"00:52.100","Text":"Since we have the molarity and the volume,"},{"Start":"00:52.100 ","End":"00:58.085","Text":"we will use the equation n moles equals m molarity times V, the volume."},{"Start":"00:58.085 ","End":"01:00.790","Text":"Notice we just multiplied by the volume both sides."},{"Start":"01:00.790 ","End":"01:05.945","Text":"This equals 0.250 from here times 80 milliliters,"},{"Start":"01:05.945 ","End":"01:07.960","Text":"which is the volume from here."},{"Start":"01:07.960 ","End":"01:10.300","Text":"First of all, we need the volume in liters."},{"Start":"01:10.300 ","End":"01:12.485","Text":"We\u0027re going to multiply by a conversion factor."},{"Start":"01:12.485 ","End":"01:15.330","Text":"In every 1 liter we have 1,000 milliliters."},{"Start":"01:15.330 ","End":"01:17.985","Text":"That way we\u0027re converting our milliliters to liters."},{"Start":"01:17.985 ","End":"01:19.730","Text":"Also instead of writing molar,"},{"Start":"01:19.730 ","End":"01:21.470","Text":"we will write moles per liter,"},{"Start":"01:21.470 ","End":"01:26.245","Text":"because molar equals moles per liter so 0.250 mole per liter"},{"Start":"01:26.245 ","End":"01:32.060","Text":"times 80 divided by 1,000 so that\u0027s 0.08 liters."},{"Start":"01:32.060 ","End":"01:33.703","Text":"Liters cancel out,"},{"Start":"01:33.703 ","End":"01:38.850","Text":"and we get 0.02 mole of glucose."},{"Start":"01:38.850 ","End":"01:40.985","Text":"Now since we want grams,"},{"Start":"01:40.985 ","End":"01:42.635","Text":"we will use the equation,"},{"Start":"01:42.635 ","End":"01:48.300","Text":"n moles equals the mass divided by the molar mass,"},{"Start":"01:48.300 ","End":"01:49.725","Text":"and we\u0027re looking for the mass."},{"Start":"01:49.725 ","End":"01:53.985","Text":"The mass equals the moles times the molar mass,"},{"Start":"01:53.985 ","End":"01:55.980","Text":"the moles we calculated,"},{"Start":"01:55.980 ","End":"01:58.275","Text":"and we still need the molar mass."},{"Start":"01:58.275 ","End":"02:00.980","Text":"The molar mass of glucose, which is here,"},{"Start":"02:00.980 ","End":"02:09.470","Text":"C_6H_12O_6 equals 6 times the molar mass of carbon since we have 6 carbons plus"},{"Start":"02:09.470 ","End":"02:14.040","Text":"12 times the molar mass of hydrogen since we have"},{"Start":"02:14.040 ","End":"02:19.245","Text":"12 hydrogens plus 6 times the molar mass of oxygen since we have 6 oxygens."},{"Start":"02:19.245 ","End":"02:26.615","Text":"This equals 6 times 12 plus 12 times 1 plus 6 times 16."},{"Start":"02:26.615 ","End":"02:30.940","Text":"This equals 180 grams per mole."},{"Start":"02:30.940 ","End":"02:34.895","Text":"The molar masses of course are taken from the periodic table of elements."},{"Start":"02:34.895 ","End":"02:36.830","Text":"As we said, to find the mass,"},{"Start":"02:36.830 ","End":"02:40.115","Text":"we\u0027re going to multiply the moles by the molar mass."},{"Start":"02:40.115 ","End":"02:42.080","Text":"The mass equals moles,"},{"Start":"02:42.080 ","End":"02:43.715","Text":"which is 0.02 mole,"},{"Start":"02:43.715 ","End":"02:45.230","Text":"times the molar mass,"},{"Start":"02:45.230 ","End":"02:47.135","Text":"which is 180 grams per mole,"},{"Start":"02:47.135 ","End":"02:48.825","Text":"moles cancel out,"},{"Start":"02:48.825 ","End":"02:56.615","Text":"and we get 3.6 grams so 3.6 grams is the mass of glucose that we get for a."},{"Start":"02:56.615 ","End":"02:58.385","Text":"That is our final answer."},{"Start":"02:58.385 ","End":"03:00.455","Text":"Thank you very much for watching."},{"Start":"03:00.455 ","End":"03:02.285","Text":"We will answer b."},{"Start":"03:02.285 ","End":"03:08.795","Text":"How much methanol density equals 0.79 L grams per mil in milliliters must be dissolved"},{"Start":"03:08.795 ","End":"03:15.530","Text":"in water to produce 3.15 liters of 0.375 molar methanol?"},{"Start":"03:15.530 ","End":"03:20.750","Text":"In b we need to find the amount of methanol in milliliters which must be dissolved"},{"Start":"03:20.750 ","End":"03:26.855","Text":"in water to produce 3.15 liters of 0.375 molar."},{"Start":"03:26.855 ","End":"03:30.110","Text":"In order to find the amount in milliliters,"},{"Start":"03:30.110 ","End":"03:31.790","Text":"we will first find the mass,"},{"Start":"03:31.790 ","End":"03:33.620","Text":"and then with the density,"},{"Start":"03:33.620 ","End":"03:35.135","Text":"we can find the milliliters."},{"Start":"03:35.135 ","End":"03:37.900","Text":"In order to find the mass, we will first use the equation,"},{"Start":"03:37.900 ","End":"03:42.365","Text":"m molarity equals n moles of solute divided by V,"},{"Start":"03:42.365 ","End":"03:45.155","Text":"which is the volume of the solution in liters."},{"Start":"03:45.155 ","End":"03:48.815","Text":"We are given the volume of the solution which is 3.15 liters,"},{"Start":"03:48.815 ","End":"03:50.480","Text":"and we\u0027re given the molarity of the solution,"},{"Start":"03:50.480 ","End":"03:52.675","Text":"which is 0.375 molar."},{"Start":"03:52.675 ","End":"03:58.115","Text":"We use the molarity and the volume which we are given in order to find the moles."},{"Start":"03:58.115 ","End":"04:01.070","Text":"If we multiply both sides of this equation by the volume,"},{"Start":"04:01.070 ","End":"04:02.645","Text":"we get that n,"},{"Start":"04:02.645 ","End":"04:07.250","Text":"the moles equals molarity times the volume and this"},{"Start":"04:07.250 ","End":"04:13.385","Text":"equals 0.375 molar times 3.15 liters,"},{"Start":"04:13.385 ","End":"04:16.985","Text":"and this equals 1.18 mole."},{"Start":"04:16.985 ","End":"04:21.380","Text":"Just recall for 1 second that molar is mole per liter,"},{"Start":"04:21.380 ","End":"04:27.890","Text":"and therefore molar times liters equals mole per liter times liters."},{"Start":"04:27.890 ","End":"04:30.770","Text":"Liters cancel out and we\u0027re left with moles."},{"Start":"04:31.300 ","End":"04:35.045","Text":"We have 1.18 mole of methanol."},{"Start":"04:35.045 ","End":"04:38.195","Text":"Next, as I said, we want to find the mass of methanol."},{"Start":"04:38.195 ","End":"04:40.280","Text":"For this purpose, we will use the equation"},{"Start":"04:40.280 ","End":"04:45.595","Text":"n moles equals m mass divided by the molar mass."},{"Start":"04:45.595 ","End":"04:48.250","Text":"If we multiply both sides by the molar mass,"},{"Start":"04:48.250 ","End":"04:49.475","Text":"we will get the equation,"},{"Start":"04:49.475 ","End":"04:52.990","Text":"mass equals the moles times the molar mass."},{"Start":"04:52.990 ","End":"04:55.019","Text":"We have the moles, they\u0027re calculated,"},{"Start":"04:55.019 ","End":"04:58.565","Text":"now we need to calculate the molar mass of methanol."},{"Start":"04:58.565 ","End":"05:03.006","Text":"The molar mass of ethanol equals the molar mass of carbon"},{"Start":"05:03.006 ","End":"05:07.760","Text":"because we have 1 carbon plus 4 times the molar mass of hydrogen,"},{"Start":"05:07.760 ","End":"05:11.790","Text":"since we have 4 hydrogens plus the molar mass of oxygen."},{"Start":"05:11.790 ","End":"05:16.530","Text":"This equals 12 plus 4 times 1 plus 16,"},{"Start":"05:16.530 ","End":"05:18.630","Text":"and this equals 32 grams per mole."},{"Start":"05:18.630 ","End":"05:21.290","Text":"All the molar masses of course are taken from"},{"Start":"05:21.290 ","End":"05:25.130","Text":"the periodic table of elements and our units are all grams per mole."},{"Start":"05:25.130 ","End":"05:28.310","Text":"As I said, the next step is to find the mass."},{"Start":"05:28.310 ","End":"05:31.910","Text":"The mass equals the moles times the molar mass,"},{"Start":"05:31.910 ","End":"05:33.920","Text":"which equals 1.18 mole,"},{"Start":"05:33.920 ","End":"05:36.320","Text":"which is here we calculated before,"},{"Start":"05:36.320 ","End":"05:38.240","Text":"times the molar mass,"},{"Start":"05:38.240 ","End":"05:47.490","Text":"32 grams per mole and this equals 37.76 grams."},{"Start":"05:47.490 ","End":"05:52.330","Text":"Our final step is to go from the mass to volume."},{"Start":"05:52.330 ","End":"05:59.615","Text":"For this purpose, we will use the equation d density equals m mass divided by V volume."},{"Start":"05:59.615 ","End":"06:01.715","Text":"Since we\u0027re looking for the volume,"},{"Start":"06:01.715 ","End":"06:06.725","Text":"we will use a simple math manipulation and volume equals mass divided by the density."},{"Start":"06:06.725 ","End":"06:08.795","Text":"The volume is what we need to calculate,"},{"Start":"06:08.795 ","End":"06:12.235","Text":"and equals mass divided by the density as we said."},{"Start":"06:12.235 ","End":"06:16.760","Text":"The mass is 37.76 grams and the density,"},{"Start":"06:16.760 ","End":"06:17.885","Text":"as you see as given here,"},{"Start":"06:17.885 ","End":"06:21.110","Text":"0.790 grams per mil."},{"Start":"06:21.110 ","End":"06:24.610","Text":"This equals 47.8 milliliters."},{"Start":"06:24.610 ","End":"06:28.790","Text":"Just remember one thing when you\u0027re dividing by a fraction as we are here,"},{"Start":"06:28.790 ","End":"06:31.300","Text":"grams divided by grams per milliliter,"},{"Start":"06:31.300 ","End":"06:34.090","Text":"it is the same as multiplying by the reciprocal of the fraction."},{"Start":"06:34.090 ","End":"06:37.220","Text":"It equals grams times milliliter per grams."},{"Start":"06:37.220 ","End":"06:40.370","Text":"Grams cancel out and we\u0027re left with milliliters."},{"Start":"06:40.370 ","End":"06:45.155","Text":"Final answer for V is 47.8 milliliters."},{"Start":"06:45.155 ","End":"06:47.910","Text":"Thank you very much for watching."}],"ID":21146},{"Watched":false,"Name":"Exercise 15","Duration":"4m 31s","ChapterTopicVideoID":20336,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.730","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.730 ","End":"00:04.800","Text":"Which has the higher concentration of sucrose?"},{"Start":"00:04.800 ","End":"00:08.715","Text":"A 42 percent sucrose solution by mass density equals 1.21"},{"Start":"00:08.715 ","End":"00:13.650","Text":"grams per mL or 1.40 molar sucrose, explain."},{"Start":"00:13.650 ","End":"00:17.790","Text":"First of all, we need to calculate the molarity of our first solution,"},{"Start":"00:17.790 ","End":"00:21.480","Text":"and then we can compare it to the second solution which is 1.40 molar,"},{"Start":"00:21.480 ","End":"00:24.645","Text":"and then we will see which has the highest concentration."},{"Start":"00:24.645 ","End":"00:28.512","Text":"We know that our first solution is 42 percent sucrose solution,"},{"Start":"00:28.512 ","End":"00:31.680","Text":"meaning that if we have 100 grams of solution,"},{"Start":"00:31.680 ","End":"00:34.515","Text":"we have 42 grams of sucrose."},{"Start":"00:34.515 ","End":"00:36.630","Text":"In order to solve this question,"},{"Start":"00:36.630 ","End":"00:39.810","Text":"we will assume that we have 100 grams of solution."},{"Start":"00:39.810 ","End":"00:43.400","Text":"As we know, molarity m equals n,"},{"Start":"00:43.400 ","End":"00:45.050","Text":"which is the moles of solute,"},{"Start":"00:45.050 ","End":"00:47.090","Text":"divided by v,"},{"Start":"00:47.090 ","End":"00:50.660","Text":"which is the volume of the solution in liters."},{"Start":"00:50.660 ","End":"00:53.240","Text":"First of all, you will find the moles of solute,"},{"Start":"00:53.240 ","End":"00:59.120","Text":"so mole equals mass divided by molar mass."},{"Start":"00:59.120 ","End":"01:01.115","Text":"We need to find the moles of sucrose,"},{"Start":"01:01.115 ","End":"01:02.210","Text":"the moles of the solute."},{"Start":"01:02.210 ","End":"01:03.470","Text":"The mass of sucrose,"},{"Start":"01:03.470 ","End":"01:06.230","Text":"we already have 42 grams of sucrose,"},{"Start":"01:06.230 ","End":"01:08.330","Text":"now we need to find the molar mass of sucrose."},{"Start":"01:08.330 ","End":"01:10.160","Text":"The molar mass of sucrose,"},{"Start":"01:10.160 ","End":"01:14.780","Text":"which is C_12 H_22 O_11,"},{"Start":"01:14.780 ","End":"01:18.455","Text":"equals the molar mass of carbon times 12."},{"Start":"01:18.455 ","End":"01:24.440","Text":"Since we have 12 carbons plus 22 times the molar mass of hydrogen."},{"Start":"01:24.440 ","End":"01:26.525","Text":"Since we have 22 hydrogens,"},{"Start":"01:26.525 ","End":"01:31.460","Text":"plus 11 times the molar mass of oxygen since we have 11 oxygens."},{"Start":"01:31.460 ","End":"01:38.940","Text":"This equals 12 times 12 plus 22 times 1 plus 11 times 16,"},{"Start":"01:38.940 ","End":"01:43.035","Text":"and this equals 342 g/mol."},{"Start":"01:43.035 ","End":"01:45.890","Text":"Now, of course, the molar masses are taken from"},{"Start":"01:45.890 ","End":"01:49.835","Text":"the periodic table of the elements and their units are grams per mole."},{"Start":"01:49.835 ","End":"01:52.100","Text":"Now that we know the molar mass,"},{"Start":"01:52.100 ","End":"01:53.870","Text":"we can move the mass, of course,"},{"Start":"01:53.870 ","End":"01:56.330","Text":"we can calculate the moles."},{"Start":"01:56.330 ","End":"01:59.885","Text":"The moles of sucrose equals the mass of sucrose,"},{"Start":"01:59.885 ","End":"02:01.370","Text":"which is 42 grams,"},{"Start":"02:01.370 ","End":"02:03.065","Text":"divided by the molar mass of sucrose,"},{"Start":"02:03.065 ","End":"02:08.610","Text":"which is 342 g/mol and this comes to 0.122 mole."},{"Start":"02:08.610 ","End":"02:10.310","Text":"We have the moles of sucrose,"},{"Start":"02:10.310 ","End":"02:11.870","Text":"which is the moles of solid,"},{"Start":"02:11.870 ","End":"02:15.050","Text":"now we need to find the volume of the solution in liters."},{"Start":"02:15.050 ","End":"02:17.210","Text":"In order to find the volume of the solution,"},{"Start":"02:17.210 ","End":"02:19.055","Text":"we will use the density of the solution,"},{"Start":"02:19.055 ","End":"02:22.290","Text":"which equals 1.21 g/ml."},{"Start":"02:22.290 ","End":"02:29.960","Text":"Right here, density equals 1.21 g per mL and recall that the density"},{"Start":"02:29.960 ","End":"02:33.860","Text":"equals the mass divided by the volume and the density"},{"Start":"02:33.860 ","End":"02:37.910","Text":"of the solution equals the mass of the solution divided by the volume of the solution."},{"Start":"02:37.910 ","End":"02:40.100","Text":"Now we need to calculate the volume of the solution,"},{"Start":"02:40.100 ","End":"02:43.120","Text":"so you can use a simple math manipulation,"},{"Start":"02:43.120 ","End":"02:48.049","Text":"and the volume equals m the mass divided by d, the density."},{"Start":"02:48.049 ","End":"02:50.600","Text":"Again v, the volume equals m,"},{"Start":"02:50.600 ","End":"02:52.805","Text":"which is our mass,"},{"Start":"02:52.805 ","End":"02:54.545","Text":"the mass of the solution."},{"Start":"02:54.545 ","End":"02:57.770","Text":"Recall that we assume that we have 100 grams of solution,"},{"Start":"02:57.770 ","End":"03:01.220","Text":"so it\u0027s 100 grams divided by the density,"},{"Start":"03:01.220 ","End":"03:04.160","Text":"which is 1.21 g/mL,"},{"Start":"03:04.160 ","End":"03:09.510","Text":"so the volume of the solution equals 82.64 Millilitres."},{"Start":"03:11.740 ","End":"03:15.050","Text":"Now that we have the moles and the volume,"},{"Start":"03:15.050 ","End":"03:16.940","Text":"we can calculate the molarity."},{"Start":"03:16.940 ","End":"03:23.675","Text":"Again, the molarity equals moles of solute divided by volume of the solution in liters."},{"Start":"03:23.675 ","End":"03:25.970","Text":"The moles of solute, which we found before,"},{"Start":"03:25.970 ","End":"03:31.340","Text":"0.122 moles divided by the volume of the solution,"},{"Start":"03:31.340 ","End":"03:35.605","Text":"which equals 82.64 millilitres."},{"Start":"03:35.605 ","End":"03:37.520","Text":"Recall we need it in liters,"},{"Start":"03:37.520 ","End":"03:40.700","Text":"so we will multiply it by a conversion factor,"},{"Start":"03:40.700 ","End":"03:44.237","Text":"1 liter for every 1,000 mLs,"},{"Start":"03:44.237 ","End":"03:46.650","Text":"the milliliters cancel out."},{"Start":"03:46.650 ","End":"03:52.200","Text":"Again, 0.122 mole divided by"},{"Start":"03:52.200 ","End":"04:00.730","Text":"0.0826 liters and this comes to 1.48 mol per liter, which equals molar."},{"Start":"04:00.730 ","End":"04:06.830","Text":"The molarity of sucrose of this first solution is found to be 1.48 molar,"},{"Start":"04:06.830 ","End":"04:09.890","Text":"and we were asked which has the higher molarity?"},{"Start":"04:09.890 ","End":"04:14.840","Text":"The first solution or the second solution which is 1.40 molar so 1.48,"},{"Start":"04:14.840 ","End":"04:18.280","Text":"of course, is greater than 1.40."},{"Start":"04:18.280 ","End":"04:22.830","Text":"Therefore the solution which was 42 percent"},{"Start":"04:22.830 ","End":"04:27.845","Text":"sucrose of first solution is the solution with the higher concentration."},{"Start":"04:27.845 ","End":"04:29.330","Text":"That is our final answer."},{"Start":"04:29.330 ","End":"04:32.010","Text":"Thank you very much for watching."}],"ID":21147},{"Watched":false,"Name":"Diluting Solutions","Duration":"6m 33s","ChapterTopicVideoID":16926,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.430","Text":"In the previous video,"},{"Start":"00:02.430 ","End":"00:04.290","Text":"we defined the molarity."},{"Start":"00:04.290 ","End":"00:07.770","Text":"The definition of molarity is that M,"},{"Start":"00:07.770 ","End":"00:10.005","Text":"the molarity, is equal to n,"},{"Start":"00:10.005 ","End":"00:12.248","Text":"the number of moles of solute,"},{"Start":"00:12.248 ","End":"00:16.575","Text":"divided by V, the volume of solution in liters."},{"Start":"00:16.575 ","End":"00:21.990","Text":"From that, we worked out that the number of moles is given by n,"},{"Start":"00:21.990 ","End":"00:25.365","Text":"the number of moles equal to the product of M,"},{"Start":"00:25.365 ","End":"00:29.580","Text":"the molarity, and V, the volume,"},{"Start":"00:29.580 ","End":"00:36.120","Text":"which we can write without the cross as M times V moles."},{"Start":"00:36.120 ","End":"00:41.760","Text":"In this video, we will learn how to dilute solutions from"},{"Start":"00:41.760 ","End":"00:47.510","Text":"a higher molarity to a lower molarity."},{"Start":"00:47.510 ","End":"00:53.765","Text":"The key to understanding dilution is to realize that we\u0027re adding more solvent,"},{"Start":"00:53.765 ","End":"00:59.120","Text":"for example, more water without changing the amount of the solute."},{"Start":"00:59.120 ","End":"01:02.510","Text":"In other words, the number of moles,"},{"Start":"01:02.510 ","End":"01:04.685","Text":"n, doesn\u0027t change."},{"Start":"01:04.685 ","End":"01:08.230","Text":"Let\u0027s write the equation that we had before."},{"Start":"01:08.230 ","End":"01:12.540","Text":"Before we try to dilute the solution,"},{"Start":"01:12.540 ","End":"01:15.970","Text":"and we\u0027ll use 1 to indicate before."},{"Start":"01:15.970 ","End":"01:22.450","Text":"N_1=M_1V_1. Now, after we\u0027ve diluted the solution,"},{"Start":"01:22.450 ","End":"01:28.520","Text":"we can write N_2=M_2V_2, that\u0027s afterwards."},{"Start":"01:28.520 ","End":"01:33.010","Text":"Now, if the number of moles doesn\u0027t change, then n_1,"},{"Start":"01:33.010 ","End":"01:34.735","Text":"the number of moles before,"},{"Start":"01:34.735 ","End":"01:37.030","Text":"is exactly equal to n_2,"},{"Start":"01:37.030 ","End":"01:38.875","Text":"the number of moles after."},{"Start":"01:38.875 ","End":"01:40.900","Text":"We conclude from that,"},{"Start":"01:40.900 ","End":"01:47.840","Text":"that M_1V_1 is equal to M_2V_2."},{"Start":"01:47.910 ","End":"01:52.479","Text":"This is the equation we\u0027re going to use for dilution."},{"Start":"01:52.479 ","End":"01:54.295","Text":"Let\u0027s take an example."},{"Start":"01:54.295 ","End":"01:59.800","Text":"Supposing we have available a 1-molar solution of hydrochloric acid,"},{"Start":"01:59.800 ","End":"02:04.070","Text":"that\u0027s HCl in water, HCl aqueous."},{"Start":"02:04.070 ","End":"02:08.150","Text":"How much of the solution do we need to dilute to produce"},{"Start":"02:08.150 ","End":"02:13.670","Text":"250 milliliters of 0.1 molar HCl."},{"Start":"02:13.670 ","End":"02:19.630","Text":"Now, the way to tackle this problem is to write out what M_1 is,"},{"Start":"02:19.630 ","End":"02:23.350","Text":"what V_1 is, what M_2 is, what V_2 is."},{"Start":"02:23.350 ","End":"02:28.245","Text":"We\u0027ll see that we know 3 of them and we have to find the fourth one."},{"Start":"02:28.245 ","End":"02:31.175","Text":"The question tells us that M_1,"},{"Start":"02:31.175 ","End":"02:32.630","Text":"the molarity at the beginning,"},{"Start":"02:32.630 ","End":"02:34.430","Text":"was 1 molar,"},{"Start":"02:34.430 ","End":"02:38.930","Text":"1 M. After dilution, it was M_2."},{"Start":"02:38.930 ","End":"02:43.070","Text":"M_2 is equal to 0.1 M. We\u0027re diluting"},{"Start":"02:43.070 ","End":"02:47.435","Text":"from a concentrated solution of 1 M to a more dilute solution of"},{"Start":"02:47.435 ","End":"02:57.260","Text":"0.1 M. It also tells us that the volume after we\u0027ve diluted it is 250 milliliters."},{"Start":"02:57.260 ","End":"03:03.860","Text":"We want to prepare 250 milliliters of diluted hydrochloric acid."},{"Start":"03:03.860 ","End":"03:10.775","Text":"Now we can write that V_2 as 0.25 liters because remember,"},{"Start":"03:10.775 ","End":"03:14.600","Text":"in all of this, the volume was in liters."},{"Start":"03:14.600 ","End":"03:17.915","Text":"What remains for us to find out is,"},{"Start":"03:17.915 ","End":"03:19.535","Text":"what is the volume of V_1,"},{"Start":"03:19.535 ","End":"03:22.195","Text":"the volume before dilution."},{"Start":"03:22.195 ","End":"03:25.770","Text":"Now, from M_1V_1=M_2V_2,"},{"Start":"03:25.840 ","End":"03:35.350","Text":"we can write that V_1 is equal to M_2 divided by M_1 times V_2."},{"Start":"03:35.350 ","End":"03:38.990","Text":"In other words, in order to find V_1,"},{"Start":"03:38.990 ","End":"03:42.430","Text":"we\u0027re just dividing the whole equation by M_1."},{"Start":"03:42.430 ","End":"03:48.175","Text":"We have M_2 divided by M_1 multiplied by V_2."},{"Start":"03:48.175 ","End":"03:51.709","Text":"All that remains is to insert the numbers."},{"Start":"03:51.709 ","End":"03:57.070","Text":"M_2 is 0.1 M,"},{"Start":"03:57.070 ","End":"04:00.410","Text":"M_1 is 1 M,"},{"Start":"04:00.410 ","End":"04:04.290","Text":"V_2 is 0.25 liters."},{"Start":"04:04.400 ","End":"04:11.430","Text":"Now, 0.1 divided by 1 is of course just 0.1."},{"Start":"04:11.430 ","End":"04:15.698","Text":"M cancels with the M,"},{"Start":"04:15.698 ","End":"04:21.615","Text":"and we\u0027re left with 0.1 times 0.25,"},{"Start":"04:21.615 ","End":"04:26.775","Text":"that\u0027s 0.025, and the units are just liters."},{"Start":"04:26.775 ","End":"04:30.260","Text":"We have 0.025 liters."},{"Start":"04:30.260 ","End":"04:33.920","Text":"If we want to write that in milliliters,"},{"Start":"04:33.920 ","End":"04:38.795","Text":"we can calculate that as 25 milliliters."},{"Start":"04:38.795 ","End":"04:43.640","Text":"The answer is 25 milliliters."},{"Start":"04:43.640 ","End":"04:47.285","Text":"We could have done it a slightly different way."},{"Start":"04:47.285 ","End":"04:52.685","Text":"We can note that the units of V_1 and V_2 are the same."},{"Start":"04:52.685 ","End":"04:57.245","Text":"It doesn\u0027t really matter whether we take milliliters or liters."},{"Start":"04:57.245 ","End":"04:58.805","Text":"How do we know that?"},{"Start":"04:58.805 ","End":"05:04.055","Text":"Because V_1=M_2/M_1, which is just a ratio,"},{"Start":"05:04.055 ","End":"05:09.980","Text":"has no units after we\u0027ve canceled the capital M and it\u0027s multiplied by V_2."},{"Start":"05:09.980 ","End":"05:13.495","Text":"V_1 and V_2 have the same units."},{"Start":"05:13.495 ","End":"05:15.405","Text":"We could have done it like this,"},{"Start":"05:15.405 ","End":"05:18.315","Text":"M_1V_1 is equal to M_2V_2."},{"Start":"05:18.315 ","End":"05:23.535","Text":"Again, V_1 is M_2 divided by M_1 times V_2,"},{"Start":"05:23.535 ","End":"05:27.915","Text":"0.1 for M_2, 1 for M_1."},{"Start":"05:27.915 ","End":"05:29.790","Text":"Remember the Ms cancel."},{"Start":"05:29.790 ","End":"05:33.605","Text":"If we take V_2 as 250 milliliters,"},{"Start":"05:33.605 ","End":"05:38.480","Text":"then we\u0027ll have 0.1 times 250 milliliters,"},{"Start":"05:38.480 ","End":"05:41.180","Text":"which is just 25 milliliters."},{"Start":"05:41.180 ","End":"05:44.135","Text":"We got precisely the same answer,"},{"Start":"05:44.135 ","End":"05:48.545","Text":"whether we take liters or milliliters."},{"Start":"05:48.545 ","End":"05:51.140","Text":"Now, how would you do this experimentally?"},{"Start":"05:51.140 ","End":"05:56.735","Text":"We would take 25 milliliters of the 1-molar solution"},{"Start":"05:56.735 ","End":"06:03.380","Text":"in a pipette or little dropper and place it in the bottom of a volumetric flask."},{"Start":"06:03.380 ","End":"06:06.860","Text":"Remember we have volumetric flask of all different volumes."},{"Start":"06:06.860 ","End":"06:11.845","Text":"Here, we need one that holds 250 milliliters."},{"Start":"06:11.845 ","End":"06:19.940","Text":"Then all we have to do is to add water right up to the 250 milliliter mark on the neck."},{"Start":"06:19.940 ","End":"06:25.820","Text":"Remember, we just have to add water right up to this mark,"},{"Start":"06:25.820 ","End":"06:28.070","Text":"and then we have our solution."},{"Start":"06:28.070 ","End":"06:33.480","Text":"In this video, we learned how to dilute a concentrated solution."}],"ID":21148},{"Watched":false,"Name":"Exercise 16","Duration":"1m 39s","ChapterTopicVideoID":20337,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.195","Text":"Hi. We are going to solve the following exercise."},{"Start":"00:03.195 ","End":"00:06.480","Text":"A 15-milliliters sample of 3.05 molar"},{"Start":"00:06.480 ","End":"00:10.290","Text":"potassium nitrate is diluted to a volume of 260 milliliters."},{"Start":"00:10.290 ","End":"00:13.365","Text":"What is the concentration of the diluted sample?"},{"Start":"00:13.365 ","End":"00:18.195","Text":"From the question, we know that the initial volume of the sample is 15 milliliters."},{"Start":"00:18.195 ","End":"00:22.965","Text":"We also know that the initial molarity of the sample is 3.05 molar."},{"Start":"00:22.965 ","End":"00:26.070","Text":"We\u0027re also given the volume of the diluted sample,"},{"Start":"00:26.070 ","End":"00:28.185","Text":"which equals 260 milliliters."},{"Start":"00:28.185 ","End":"00:32.470","Text":"The final volume equals 260 milliliters."},{"Start":"00:33.050 ","End":"00:37.320","Text":"We\u0027re asked to find the concentration of the diluted sample."},{"Start":"00:37.320 ","End":"00:40.020","Text":"For this purpose, we will use the equation,"},{"Start":"00:40.020 ","End":"00:48.820","Text":"initial molarity times initial volume equals final molarity times final volume."},{"Start":"00:48.820 ","End":"00:51.735","Text":"We\u0027re looking for the final molarity."},{"Start":"00:51.735 ","End":"01:00.200","Text":"The final molarity equals the initial molarity times the initial volume,"},{"Start":"01:00.200 ","End":"01:03.570","Text":"divided by the final volume."},{"Start":"01:05.150 ","End":"01:11.090","Text":"The initial molarity equals 3.05 molar,"},{"Start":"01:11.090 ","End":"01:13.610","Text":"times 15 milliliters,"},{"Start":"01:13.610 ","End":"01:15.470","Text":"which is the initial volume,"},{"Start":"01:15.470 ","End":"01:19.550","Text":"divided by the final volume,"},{"Start":"01:19.550 ","End":"01:22.200","Text":"which equals 260 milliliters."},{"Start":"01:23.090 ","End":"01:29.105","Text":"The milliliters cancel out and we receive 0.18 molar."},{"Start":"01:29.105 ","End":"01:35.015","Text":"The final molarity of the diluted sample equals 0.18 molar."},{"Start":"01:35.015 ","End":"01:36.905","Text":"That is our final answer."},{"Start":"01:36.905 ","End":"01:39.420","Text":"Thank you very much for watching."}],"ID":21149},{"Watched":false,"Name":"Exercise 17","Duration":"2m 23s","ChapterTopicVideoID":20338,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.225","Text":"Hi. We are going to answer the following exercise."},{"Start":"00:03.225 ","End":"00:08.535","Text":"A 30 milliliter sample of HCl is diluted to a volume of 700 ml."},{"Start":"00:08.535 ","End":"00:15.405","Text":"If the concentration of the diluted sample is 0.080 molar HCl,"},{"Start":"00:15.405 ","End":"00:18.375","Text":"what was the concentration of the original solution?"},{"Start":"00:18.375 ","End":"00:21.995","Text":"We have the original solution which we\u0027re going to call the initial solution,"},{"Start":"00:21.995 ","End":"00:25.550","Text":"and we note that the volume here is 30 ml."},{"Start":"00:25.550 ","End":"00:29.105","Text":"The volume of the initial solution equals 30 ml."},{"Start":"00:29.105 ","End":"00:33.140","Text":"We also know that the volume of the diluted solution is 700 ml,"},{"Start":"00:33.140 ","End":"00:34.820","Text":"and we\u0027re going to call this the volume of"},{"Start":"00:34.820 ","End":"00:39.170","Text":"the final solution after the solution has been diluted."},{"Start":"00:39.170 ","End":"00:46.445","Text":"We also know the concentration here of the diluted sample is 0.080 molar HCl."},{"Start":"00:46.445 ","End":"00:50.630","Text":"The concentration M final of the diluted sample,"},{"Start":"00:50.630 ","End":"00:56.390","Text":"which is the final sample, is 0.080 molar."},{"Start":"00:56.390 ","End":"00:59.150","Text":"We\u0027re asked to find the concentration of the original solution,"},{"Start":"00:59.150 ","End":"01:02.640","Text":"meaning the initial concentration."},{"Start":"01:02.640 ","End":"01:04.795","Text":"In order to solve this question,"},{"Start":"01:04.795 ","End":"01:07.600","Text":"we\u0027re going to use the equation M_i,"},{"Start":"01:07.600 ","End":"01:10.540","Text":"meaning the initial concentration times V_i,"},{"Start":"01:10.540 ","End":"01:12.870","Text":"the initial volume equals M_f,"},{"Start":"01:12.870 ","End":"01:17.210","Text":"final concentration times V_f, final volume."},{"Start":"01:17.210 ","End":"01:19.615","Text":"Since we\u0027re looking for the initial concentration,"},{"Start":"01:19.615 ","End":"01:22.895","Text":"we will divide both sides by the initial volume."},{"Start":"01:22.895 ","End":"01:31.530","Text":"The initial concentration equals M_f times V_f divided by V_i."},{"Start":"01:31.530 ","End":"01:34.110","Text":"As you can see, all these are given."},{"Start":"01:34.110 ","End":"01:37.975","Text":"M_f equals 0.080 molar,"},{"Start":"01:37.975 ","End":"01:40.330","Text":"times V_f, which is 700 ml,"},{"Start":"01:40.330 ","End":"01:43.790","Text":"divided by V_i, which is 30 ml."},{"Start":"01:44.300 ","End":"01:48.290","Text":"This equals 1.87 molar."},{"Start":"01:48.290 ","End":"01:53.900","Text":"Now it\u0027s important for me just to mention that usually when we\u0027re speaking of molarity,"},{"Start":"01:53.900 ","End":"01:59.022","Text":"and we have molarity equals moles of solute per volume of solution,"},{"Start":"01:59.022 ","End":"02:01.970","Text":"the volume of the solution needs to be in liters."},{"Start":"02:01.970 ","End":"02:07.565","Text":"But since here we have volume on both sides of the equation,"},{"Start":"02:07.565 ","End":"02:10.610","Text":"we can use any units for the volume, for example,"},{"Start":"02:10.610 ","End":"02:15.095","Text":"milliliters, as long as we have the same units on both sides."},{"Start":"02:15.095 ","End":"02:18.220","Text":"That\u0027s what we used here. We used milliliters on both sides."},{"Start":"02:18.220 ","End":"02:21.155","Text":"This is our final answer for this question."},{"Start":"02:21.155 ","End":"02:23.670","Text":"Thank you very much for watching."}],"ID":21150},{"Watched":false,"Name":"Exercise 18","Duration":"4m 32s","ChapterTopicVideoID":22950,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"Here, we going to solve the following exercise."},{"Start":"00:02.790 ","End":"00:09.330","Text":"Two silver nitrate plus sodium sulfide gives us silver sulfide plus 2 sodium nitrate."},{"Start":"00:09.330 ","End":"00:13.560","Text":"How many grams of sodium sulfide are required to react completely"},{"Start":"00:13.560 ","End":"00:17.790","Text":"with 35 milliliters of 0.145 molar silver nitrate?"},{"Start":"00:17.790 ","End":"00:23.770","Text":"From the question, we know that the volume of the silver nitrate is 35 milliliters."},{"Start":"00:28.400 ","End":"00:37.959","Text":"We all insinuate that the molarity of the silver nitrate is 0.145 molar."},{"Start":"00:37.959 ","End":"00:44.990","Text":"We\u0027re asked to find the grams of sodium sulfide."},{"Start":"00:44.990 ","End":"00:47.705","Text":"First, we\u0027ll find the moles of sodium sulfide,"},{"Start":"00:47.705 ","End":"00:50.000","Text":"and to find the moles of sodium sulfide,"},{"Start":"00:50.000 ","End":"00:52.625","Text":"we\u0027ll first find the moles of the silver nitrate."},{"Start":"00:52.625 ","End":"00:55.880","Text":"Recall that M molarity equals n,"},{"Start":"00:55.880 ","End":"00:59.525","Text":"the number of moles of solute divided by V,"},{"Start":"00:59.525 ","End":"01:01.280","Text":"which is the volume of the solution."},{"Start":"01:01.280 ","End":"01:03.920","Text":"If we multiply both sides by the volume,"},{"Start":"01:03.920 ","End":"01:11.000","Text":"we get n moles of the silver nitrate equals M,"},{"Start":"01:11.000 ","End":"01:18.600","Text":"molarity of silver nitrate times the volume of silver nitrate."},{"Start":"01:20.080 ","End":"01:27.175","Text":"This equals 0.145 molar times 35 mil."},{"Start":"01:27.175 ","End":"01:30.470","Text":"But recall that we need the volume in liters,"},{"Start":"01:30.470 ","End":"01:33.080","Text":"therefore, we multiply it by a conversion factor."},{"Start":"01:33.080 ","End":"01:36.120","Text":"In every 1 liter, we have 1,000 milliliters,"},{"Start":"01:36.120 ","End":"01:37.920","Text":"the milliliters cancel out."},{"Start":"01:37.920 ","End":"01:39.395","Text":"Here, we are left with liter,"},{"Start":"01:39.395 ","End":"01:44.250","Text":"and also molar is mole per liter,"},{"Start":"01:44.250 ","End":"01:53.440","Text":"so the liters cancel out and we get 5.08 times 10 to the negative 3 mole,"},{"Start":"01:53.440 ","End":"01:55.640","Text":"which is the same as 5.08 millimole."},{"Start":"01:55.640 ","End":"01:59.030","Text":"In order to find the moles of the sodium sulfide,"},{"Start":"01:59.030 ","End":"02:02.015","Text":"we will use the moles of the silver nitrate,"},{"Start":"02:02.015 ","End":"02:03.500","Text":"which we calculated,"},{"Start":"02:03.500 ","End":"02:06.950","Text":"and if we look at our equation for every 2 moles of silver nitrate,"},{"Start":"02:06.950 ","End":"02:09.680","Text":"we get 1 mole of sodium sulfide."},{"Start":"02:09.680 ","End":"02:20.460","Text":"So we multiply this by 1 mole of sodium sulfide for every 2 moles of silver nitrate."},{"Start":"02:20.460 ","End":"02:23.505","Text":"The moles of silver nitrate are in the denominator,"},{"Start":"02:23.505 ","End":"02:25.880","Text":"that way the moles of silver nitrate will cancel out."},{"Start":"02:25.880 ","End":"02:28.925","Text":"We\u0027ll be left with the moles of the sodium sulfide,"},{"Start":"02:28.925 ","End":"02:37.400","Text":"so this equals 5.08 times 10 to the negative 3 mole of silver nitrate"},{"Start":"02:37.400 ","End":"02:48.260","Text":"times 1 mole of sodium sulfide for every 2 moles of silver nitrate."},{"Start":"02:48.260 ","End":"02:53.580","Text":"This equals 2.54 times 10 to"},{"Start":"02:53.580 ","End":"02:59.780","Text":"the negative 3 moles of sodium sulfide."},{"Start":"02:59.780 ","End":"03:05.465","Text":"Now in the next step, we want to turn the moles of sodium sulfide into grams."},{"Start":"03:05.465 ","End":"03:09.290","Text":"For this purpose, we will use the equation n,"},{"Start":"03:09.290 ","End":"03:10.895","Text":"the number of moles equals M,"},{"Start":"03:10.895 ","End":"03:15.360","Text":"which is the mass divided by the molar mass."},{"Start":"03:15.360 ","End":"03:17.670","Text":"The molar mass,"},{"Start":"03:17.670 ","End":"03:23.050","Text":"if sodium sulfide equals 2 times the molar mass of"},{"Start":"03:23.050 ","End":"03:30.285","Text":"sodium plus 1 times the molar mass of sulfur."},{"Start":"03:30.285 ","End":"03:35.875","Text":"This equals 2 times 23 plus 32.07."},{"Start":"03:35.875 ","End":"03:37.090","Text":"The molar masses, of course,"},{"Start":"03:37.090 ","End":"03:39.535","Text":"are taken from the periodic table of elements,"},{"Start":"03:39.535 ","End":"03:45.970","Text":"and this equals 78.07 g/mol."},{"Start":"03:45.970 ","End":"03:49.060","Text":"Now we want to find the mass of the sodium sulfide."},{"Start":"03:49.060 ","End":"03:51.880","Text":"For that purpose, we will multiply both sides by the molar mass,"},{"Start":"03:51.880 ","End":"03:55.150","Text":"so the mass of sodium sulfide equals"},{"Start":"03:55.150 ","End":"04:00.340","Text":"the moles of sodium sulfide times the molar mass of sodium sulfide."},{"Start":"04:00.340 ","End":"04:04.115","Text":"The moles of sodium sulfide are given here and we calculated them,"},{"Start":"04:04.115 ","End":"04:12.665","Text":"2.54 times 10 to the negative 3-mole times the molar mass,"},{"Start":"04:12.665 ","End":"04:17.760","Text":"which we also calculated and equals 78.07 g/mol."},{"Start":"04:20.440 ","End":"04:27.514","Text":"The moles cancel out and we get 0.2 grams of sodium sulfide."},{"Start":"04:27.514 ","End":"04:30.065","Text":"That is our final answer."},{"Start":"04:30.065 ","End":"04:33.120","Text":"Thank you very much for watching."}],"ID":23796},{"Watched":false,"Name":"Exercise 19","Duration":"5m 11s","ChapterTopicVideoID":22951,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.865","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.865 ","End":"00:05.880","Text":"How many milliliters of 0.560 molar"},{"Start":"00:05.880 ","End":"00:09.510","Text":"potassium chromate are needed to precipitate all of the silver ion"},{"Start":"00:09.510 ","End":"00:16.740","Text":"in 385 milliliters of 0.156 molar silver nitrate as silver chromate."},{"Start":"00:16.740 ","End":"00:19.080","Text":"2 silver nitrate plus"},{"Start":"00:19.080 ","End":"00:24.150","Text":"potassium chromate gives us silver chromate plus 2 potassium nitrate."},{"Start":"00:24.150 ","End":"00:28.245","Text":"In this question, we\u0027re asked to find the volume of potassium chromate."},{"Start":"00:28.245 ","End":"00:30.615","Text":"In order to find the volume of potassium chromate,"},{"Start":"00:30.615 ","End":"00:33.945","Text":"we will first find the moles of the silver nitrate,"},{"Start":"00:33.945 ","End":"00:37.935","Text":"then calculate the moles of potassium chromate,"},{"Start":"00:37.935 ","End":"00:40.320","Text":"and then find the volume of potassium chromate."},{"Start":"00:40.320 ","End":"00:43.505","Text":"In order to find the moles of the silver nitrate,"},{"Start":"00:43.505 ","End":"00:48.335","Text":"we will use the equation M molarity equals n,"},{"Start":"00:48.335 ","End":"00:51.350","Text":"which is the moles of solute divide by V,"},{"Start":"00:51.350 ","End":"00:54.140","Text":"which is the volume of the solution in liters."},{"Start":"00:54.140 ","End":"00:57.010","Text":"We\u0027re looking for the moles of silver nitrate,"},{"Start":"00:57.010 ","End":"00:59.810","Text":"therefore, we will multiply both sides by the volume."},{"Start":"00:59.810 ","End":"01:03.290","Text":"The molar silver nitrate equal"},{"Start":"01:03.290 ","End":"01:08.130","Text":"the molarity of silver nitrate times the volume of the silver nitrate,"},{"Start":"01:11.150 ","End":"01:14.330","Text":"and this equals the molarity of the silver nitrate,"},{"Start":"01:14.330 ","End":"01:16.715","Text":"which is 0.156 molar,"},{"Start":"01:16.715 ","End":"01:20.370","Text":"times the volume which is 385 milliliters."},{"Start":"01:21.760 ","End":"01:25.010","Text":"But remember that we need the volume in liters."},{"Start":"01:25.010 ","End":"01:27.320","Text":"Therefore, we will multiply by a conversion factor,"},{"Start":"01:27.320 ","End":"01:30.170","Text":"and 1 liter, we have 1,000 milliliters."},{"Start":"01:30.170 ","End":"01:31.865","Text":"The milliliters cancel out,"},{"Start":"01:31.865 ","End":"01:33.545","Text":"and here we are left with liter,"},{"Start":"01:33.545 ","End":"01:39.845","Text":"and also molarity or molar equals mole per liter."},{"Start":"01:39.845 ","End":"01:47.650","Text":"The liters cancel out and our units will be mole so this comes to 0.06 mole."},{"Start":"01:47.650 ","End":"01:51.395","Text":"The moles of the silver nitrate are 0.06 mole."},{"Start":"01:51.395 ","End":"01:53.990","Text":"Now we will calculate the moles of potassium chromate."},{"Start":"01:53.990 ","End":"01:57.285","Text":"The moles of potassium chromate equal the moles of"},{"Start":"01:57.285 ","End":"02:01.550","Text":"silver nitrate times if we look at our equation,"},{"Start":"02:01.550 ","End":"02:04.130","Text":"we can see that for every 2 moles of silver nitrate,"},{"Start":"02:04.130 ","End":"02:07.310","Text":"which reacts, 1 mole of potassium chromate reacts."},{"Start":"02:07.310 ","End":"02:17.310","Text":"We multiply this by 1 mole of potassium chromate divided by 2 moles of silver nitrate."},{"Start":"02:18.140 ","End":"02:21.080","Text":"The silver nitrate is in the denominator,"},{"Start":"02:21.080 ","End":"02:23.630","Text":"since we\u0027re multiplying by moles of silver nitrate,"},{"Start":"02:23.630 ","End":"02:26.203","Text":"that way, the moles of silver nitrate will cancel out,"},{"Start":"02:26.203 ","End":"02:35.430","Text":"and we will be left with moles of potassium chromate so this equals 0.06 mole of"},{"Start":"02:35.430 ","End":"02:40.830","Text":"silver nitrate times 1 mole"},{"Start":"02:40.830 ","End":"02:50.430","Text":"potassium chromate divided by 2 moles of silver nitrate."},{"Start":"02:50.430 ","End":"02:56.530","Text":"The moles of silver nitrate cancel out and we get 0.03 mole."},{"Start":"02:56.530 ","End":"03:00.925","Text":"The moles of the potassium chromate, are 0.03 mole."},{"Start":"03:00.925 ","End":"03:04.555","Text":"Now we will find the volume of potassium chromate."},{"Start":"03:04.555 ","End":"03:06.550","Text":"In order to find the volume, again,"},{"Start":"03:06.550 ","End":"03:08.350","Text":"we will use the equation,"},{"Start":"03:08.350 ","End":"03:10.345","Text":"M molarity equals n,"},{"Start":"03:10.345 ","End":"03:12.380","Text":"the number of moles of solute divided by V,"},{"Start":"03:12.380 ","End":"03:13.525","Text":"and in this case,"},{"Start":"03:13.525 ","End":"03:14.995","Text":"we are looking for the volume."},{"Start":"03:14.995 ","End":"03:18.965","Text":"Therefore, simple math manipulation will give us V,"},{"Start":"03:18.965 ","End":"03:21.620","Text":"the volume equals n,"},{"Start":"03:21.620 ","End":"03:24.589","Text":"which is the moles divided by the molarity."},{"Start":"03:24.589 ","End":"03:26.660","Text":"Of course, we\u0027re looking for the potassium chromate."},{"Start":"03:26.660 ","End":"03:30.050","Text":"Therefore, it\u0027s the volume of potassium chromate equals the moles of"},{"Start":"03:30.050 ","End":"03:36.320","Text":"potassium chromate divided by the molarity of potassium chromate."},{"Start":"03:36.320 ","End":"03:39.200","Text":"We calculated the moles of potassium chromate,"},{"Start":"03:39.200 ","End":"03:41.450","Text":"and this equals 0.03 mole"},{"Start":"03:41.450 ","End":"03:46.085","Text":"potassium chromate divided by the molarity of potassium chromate."},{"Start":"03:46.085 ","End":"03:51.710","Text":"So the molarity is given and it\u0027s 0.560 molar and molar again,"},{"Start":"03:51.710 ","End":"03:53.450","Text":"is mole per liter,"},{"Start":"03:53.450 ","End":"03:56.790","Text":"of course mole of potassium chromate."},{"Start":"03:57.770 ","End":"04:08.185","Text":"The moles cancel out and we get 0.054 liters of potassium chromate."},{"Start":"04:08.185 ","End":"04:13.030","Text":"Now I just want to remind you here that when you divide by a fraction,"},{"Start":"04:13.030 ","End":"04:17.185","Text":"It is the same as multiplying by the reciprocal of the fraction."},{"Start":"04:17.185 ","End":"04:21.220","Text":"Therefore, mole divided by a fraction,"},{"Start":"04:21.220 ","End":"04:27.820","Text":"mole per liter equals mole times liter per mole."},{"Start":"04:27.820 ","End":"04:30.175","Text":"The moles cancel out,"},{"Start":"04:30.175 ","End":"04:32.710","Text":"and we\u0027re left with liters for the unit."},{"Start":"04:32.710 ","End":"04:37.535","Text":"Now we were asked to find the volume of the potassium chromate in milliliters."},{"Start":"04:37.535 ","End":"04:44.885","Text":"Again, the volume of potassium chromate equals 0.054 liters."},{"Start":"04:44.885 ","End":"04:47.735","Text":"However, we were asked to find the volume in milliliters."},{"Start":"04:47.735 ","End":"04:51.190","Text":"For this purpose, we will multiply by a conversion factor."},{"Start":"04:51.190 ","End":"04:55.785","Text":"There are 1,000 milliliters in every 1 liter."},{"Start":"04:55.785 ","End":"04:57.585","Text":"The liters cancel out,"},{"Start":"04:57.585 ","End":"05:03.105","Text":"and our answer equals 54 milliliters."},{"Start":"05:03.105 ","End":"05:07.800","Text":"The volume of the potassium chromate equals 54 milliliters,"},{"Start":"05:07.800 ","End":"05:08.960","Text":"that is our final answer."},{"Start":"05:08.960 ","End":"05:11.490","Text":"Thank you very much for watching."}],"ID":23797},{"Watched":false,"Name":"Exercise 20","Duration":"3m 33s","ChapterTopicVideoID":22949,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.060 ","End":"00:07.140","Text":"How many grams of sodium must react with 200 milliliters of water to"},{"Start":"00:07.140 ","End":"00:11.505","Text":"produce a solution that is 0.160 molar sodium hydroxide?"},{"Start":"00:11.505 ","End":"00:15.390","Text":"Assume a final solution volume of 200 milliliters."},{"Start":"00:15.390 ","End":"00:20.040","Text":"2 sodium plus 2 water gives us 2 sodium hydroxide plus hydrogen."},{"Start":"00:20.040 ","End":"00:22.125","Text":"In order to find the grams of sodium,"},{"Start":"00:22.125 ","End":"00:24.674","Text":"we\u0027ll first find the moles of sodium hydroxide,"},{"Start":"00:24.674 ","End":"00:25.860","Text":"then moles of sodium,"},{"Start":"00:25.860 ","End":"00:27.975","Text":"and then we will convert these moles into grams."},{"Start":"00:27.975 ","End":"00:30.300","Text":"In order to find the moles of sodium hydroxide,"},{"Start":"00:30.300 ","End":"00:33.730","Text":"we will use the equation M molarity equals n,"},{"Start":"00:33.730 ","End":"00:36.605","Text":"which is the moles of solute divided by V,"},{"Start":"00:36.605 ","End":"00:39.200","Text":"which is the volume of the solution in liters."},{"Start":"00:39.200 ","End":"00:42.005","Text":"The molarity of the sodium hydroxide is given"},{"Start":"00:42.005 ","End":"00:48.390","Text":"n=0.160 molar and the volume of the solution is also given n=200 milliliters."},{"Start":"00:51.940 ","End":"00:56.645","Text":"Using the molarity and the volume of the solution,"},{"Start":"00:56.645 ","End":"01:00.590","Text":"we can find the moles of the sodium hydroxide solute."},{"Start":"01:00.590 ","End":"01:07.310","Text":"The number of moles equals the molarity times the volume."},{"Start":"01:07.310 ","End":"01:11.180","Text":"We just multiplied both sides of the equation by the volume."},{"Start":"01:11.180 ","End":"01:12.680","Text":"This equals the molarity,"},{"Start":"01:12.680 ","End":"01:18.040","Text":"which is 0.160 times the volume of the solution,"},{"Start":"01:18.040 ","End":"01:20.215","Text":"which is 200 milliliters."},{"Start":"01:20.215 ","End":"01:23.290","Text":"But recall that we need the volume in liters."},{"Start":"01:23.290 ","End":"01:25.885","Text":"Therefore, we will multiply by a conversion factor."},{"Start":"01:25.885 ","End":"01:28.525","Text":"For every 1 liter, we have 1,000 milliliters."},{"Start":"01:28.525 ","End":"01:30.220","Text":"The milliliters cancel out."},{"Start":"01:30.220 ","End":"01:34.270","Text":"This equals again 0.160 molar,"},{"Start":"01:34.270 ","End":"01:39.865","Text":"but molar equals mole per liter times 0.2 liters."},{"Start":"01:39.865 ","End":"01:47.050","Text":"Liters cancel out and we get 0.032 mole of sodium hydroxide."},{"Start":"01:47.050 ","End":"01:50.575","Text":"Now we will find the moles of sodium."},{"Start":"01:50.575 ","End":"01:57.014","Text":"This equals the moles of sodium hydroxide times,"},{"Start":"01:57.014 ","End":"01:58.875","Text":"if we look at the equation,"},{"Start":"01:58.875 ","End":"02:00.810","Text":"for every 2 moles of sodium,"},{"Start":"02:00.810 ","End":"02:03.735","Text":"we have 2 moles of sodium hydroxide."},{"Start":"02:03.735 ","End":"02:13.035","Text":"Times 2 moles of sodium divided by 2 moles of sodium hydroxide."},{"Start":"02:13.035 ","End":"02:15.235","Text":"The 2 will cancel out."},{"Start":"02:15.235 ","End":"02:17.500","Text":"The moles of sodium hydroxide,"},{"Start":"02:17.500 ","End":"02:23.560","Text":"we calculated equals 0.032 mole of"},{"Start":"02:23.560 ","End":"02:33.850","Text":"sodium hydroxide times 1 mole of sodium divided by 1 mole of sodium hydroxide."},{"Start":"02:33.850 ","End":"02:44.590","Text":"The moles of sodium hydroxide cancel out and we get 0.032 moles of sodium."},{"Start":"02:44.590 ","End":"02:47.680","Text":"Now you want to find the grams of sodium."},{"Start":"02:47.680 ","End":"02:51.130","Text":"For this purpose, we will use the equation and the number of"},{"Start":"02:51.130 ","End":"02:54.850","Text":"moles equal mass divided by molar mass."},{"Start":"02:54.850 ","End":"02:57.250","Text":"The mass, which is what we\u0027re looking for,"},{"Start":"02:57.250 ","End":"03:01.970","Text":"equals moles times the molar mass."},{"Start":"03:02.750 ","End":"03:11.170","Text":"The moles of sodium are 0.032 mole times the molar mass of sodium,"},{"Start":"03:11.170 ","End":"03:15.400","Text":"which equals 22.99 gram per mole."},{"Start":"03:15.400 ","End":"03:18.490","Text":"This value is taken from the periodic table of elements,"},{"Start":"03:18.490 ","End":"03:25.920","Text":"so the moles cancel out and we get 0.736 grams."},{"Start":"03:25.920 ","End":"03:29.660","Text":"The mass of sodium equals 0.736 grams."},{"Start":"03:29.660 ","End":"03:31.235","Text":"That is our final answer."},{"Start":"03:31.235 ","End":"03:33.660","Text":"Thank you very much for watching."}],"ID":23795},{"Watched":false,"Name":"Exercise 21","Duration":"6m ","ChapterTopicVideoID":22952,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.565","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.565 ","End":"00:08.670","Text":"A 0.4261 grams sample of oxalic acid H_2C_2O_4 requires"},{"Start":"00:08.670 ","End":"00:12.300","Text":"35.45 milliliters of a particular concentration of"},{"Start":"00:12.300 ","End":"00:16.530","Text":"sodium hydroxide aqueous solution to complete the following reaction."},{"Start":"00:16.530 ","End":"00:19.425","Text":"What is the molarity of the sodium hydroxide?"},{"Start":"00:19.425 ","End":"00:25.395","Text":"Oxalic acid plus 2 sodium hydroxide gives us sodium oxalate plus 2 water."},{"Start":"00:25.395 ","End":"00:29.040","Text":"We\u0027re asked to find the molarity of the sodium hydroxide."},{"Start":"00:29.040 ","End":"00:32.849","Text":"In order to find the molarity of the sodium hydroxide solution,"},{"Start":"00:32.849 ","End":"00:35.550","Text":"we will first find the number of moles of oxalic acid"},{"Start":"00:35.550 ","End":"00:39.255","Text":"and then calculate the number of moles of the sodium hydroxide."},{"Start":"00:39.255 ","End":"00:41.340","Text":"Then calculate the molarity."},{"Start":"00:41.340 ","End":"00:43.820","Text":"In order to find the moles of oxalic acid,"},{"Start":"00:43.820 ","End":"00:45.515","Text":"we will use the equation, n,"},{"Start":"00:45.515 ","End":"00:47.345","Text":"mole equals m,"},{"Start":"00:47.345 ","End":"00:49.850","Text":"mass divided by the molar mass."},{"Start":"00:49.850 ","End":"00:59.640","Text":"The mass of the oxalic acid is given in equals of 0.4261 grams."},{"Start":"00:59.640 ","End":"01:02.180","Text":"We will calculate the molar mass of oxalic acid."},{"Start":"01:02.180 ","End":"01:04.400","Text":"The molar mass of oxalic acid,"},{"Start":"01:04.400 ","End":"01:09.845","Text":"which is H_2C_2O_4 equals 2 times the molar mass of hydrogen,"},{"Start":"01:09.845 ","End":"01:11.870","Text":"since we have 2 hydrogens,"},{"Start":"01:11.870 ","End":"01:16.595","Text":"plus 2 times the molar mass of carbon since we have 2 carbons,"},{"Start":"01:16.595 ","End":"01:21.515","Text":"plus 4 times the molar mass of oxygen since we have 4 oxygens."},{"Start":"01:21.515 ","End":"01:28.400","Text":"This equals 2 times 1.01 gram per mole plus 2"},{"Start":"01:28.400 ","End":"01:36.935","Text":"times 12.01 grams per mole plus 4 times 16 grams per mole."},{"Start":"01:36.935 ","End":"01:40.970","Text":"The molar masses were all taken from the periodic table of elements,"},{"Start":"01:40.970 ","End":"01:48.670","Text":"and this equals 90.04 grams per mole."},{"Start":"01:48.670 ","End":"01:51.379","Text":"Now, I\u0027m going to rotate the screen down."},{"Start":"01:51.379 ","End":"01:52.490","Text":"I\u0027m just writing down,"},{"Start":"01:52.490 ","End":"01:55.010","Text":"so we don\u0027t forget that the volume of"},{"Start":"01:55.010 ","End":"01:59.970","Text":"the sodium hydroxide solution equals 35.45 milliliters."},{"Start":"02:09.460 ","End":"02:14.869","Text":"Now that we know the mass of oxalic acid and the molar mass of the oxalic acid,"},{"Start":"02:14.869 ","End":"02:16.490","Text":"we can calculate the moles."},{"Start":"02:16.490 ","End":"02:19.490","Text":"The moles of the oxalic acid equals the mass of"},{"Start":"02:19.490 ","End":"02:25.110","Text":"the oxalic acid divided by the molar mass of oxalic acid,"},{"Start":"02:25.110 ","End":"02:30.560","Text":"which equals 0.4261, which was given in the question,"},{"Start":"02:30.560 ","End":"02:37.450","Text":"divide by 90.04 grams per mole, which we calculated."},{"Start":"02:37.550 ","End":"02:43.960","Text":"This equals 4.73 times 10 to the negative 3 mole of oxalic acid."},{"Start":"02:43.960 ","End":"02:46.460","Text":"Now that we have the moles of oxalic acid,"},{"Start":"02:46.460 ","End":"02:48.770","Text":"we can calculate the moles of the sodium hydroxide."},{"Start":"02:48.770 ","End":"02:57.555","Text":"The moles of sodium hydroxide equals the moles of oxalic acid times,"},{"Start":"02:57.555 ","End":"02:58.850","Text":"if we look at our equation,"},{"Start":"02:58.850 ","End":"03:02.075","Text":"we can see that for every mole of oxalic acid that reacts,"},{"Start":"03:02.075 ","End":"03:04.925","Text":"we have 2 moles of the sodium hydroxide which reacts."},{"Start":"03:04.925 ","End":"03:15.625","Text":"This is times 2 moles of sodium hydroxide for every 1 mole of oxalic acid."},{"Start":"03:15.625 ","End":"03:18.320","Text":"The moles of oxalic acid are in the denominator."},{"Start":"03:18.320 ","End":"03:20.840","Text":"This way, the moles of oxalic acid will cancel"},{"Start":"03:20.840 ","End":"03:23.615","Text":"out and we will be left with the moles of sodium hydroxide."},{"Start":"03:23.615 ","End":"03:28.830","Text":"This equals 4.73 times 10 to the negative 3 moles of"},{"Start":"03:28.830 ","End":"03:33.660","Text":"oxalic acid times 2 moles of"},{"Start":"03:33.660 ","End":"03:42.025","Text":"sodium hydroxide divided by 1 mole of oxalic acid."},{"Start":"03:42.025 ","End":"03:45.980","Text":"The moles of oxalic acid cancel out,"},{"Start":"03:45.980 ","End":"03:56.920","Text":"and we get 9.46 times 10 to the negative 3 moles of sodium hydroxide."},{"Start":"03:57.250 ","End":"04:01.745","Text":"Now our last step is to calculate the molarity of the sodium hydroxide."},{"Start":"04:01.745 ","End":"04:03.935","Text":"We already have the number of moles,"},{"Start":"04:03.935 ","End":"04:05.645","Text":"which we just calculated,"},{"Start":"04:05.645 ","End":"04:07.130","Text":"and we have our volume."},{"Start":"04:07.130 ","End":"04:09.815","Text":"I\u0027m going to write the volume here again on this side."},{"Start":"04:09.815 ","End":"04:17.990","Text":"The volume of the sodium hydroxide solution equals 35.45 milliliters."},{"Start":"04:17.990 ","End":"04:20.450","Text":"Now we will use the equation M,"},{"Start":"04:20.450 ","End":"04:21.935","Text":"molarity equals n,"},{"Start":"04:21.935 ","End":"04:27.680","Text":"number of moles of the solute divided by V,"},{"Start":"04:27.680 ","End":"04:29.840","Text":"which is the volume of the solution in liters."},{"Start":"04:29.840 ","End":"04:35.420","Text":"To calculate the molarity of the sodium hydroxide,"},{"Start":"04:36.950 ","End":"04:39.020","Text":"we have the moles of"},{"Start":"04:39.020 ","End":"04:48.030","Text":"sodium hydroxide divided by the volume of the sodium hydroxide solution."},{"Start":"04:48.460 ","End":"04:51.650","Text":"This equals moles we calculated,"},{"Start":"04:51.650 ","End":"04:57.870","Text":"and there are 9.46 times 10 to the negative 3 moles of"},{"Start":"04:57.870 ","End":"05:05.135","Text":"sodium hydroxide divided by the volume of the sodium hydroxide solution,"},{"Start":"05:05.135 ","End":"05:08.550","Text":"which is 35.45 milliliters."},{"Start":"05:10.310 ","End":"05:13.880","Text":"Don\u0027t forget the volume needs to be in liters."},{"Start":"05:13.880 ","End":"05:16.520","Text":"Therefore, we will multiply it by the conversion factor,"},{"Start":"05:16.520 ","End":"05:19.075","Text":"1 liter for every 1,000 milliliters."},{"Start":"05:19.075 ","End":"05:24.840","Text":"The milliliters cancel out and we\u0027re left with mole per liter."},{"Start":"05:24.840 ","End":"05:33.135","Text":"This equals 0.267 mole per liter."},{"Start":"05:33.135 ","End":"05:36.740","Text":"But remember the mole per liter is molar."},{"Start":"05:36.740 ","End":"05:40.460","Text":"Therefore, the molarity of"},{"Start":"05:40.460 ","End":"05:45.305","Text":"the sodium hydroxide equals"},{"Start":"05:45.305 ","End":"05:50.900","Text":"0.267 mole per liter, which equals molar."},{"Start":"05:50.900 ","End":"05:56.585","Text":"So the molarity of the sodium hydroxide equals 0.267 molar."},{"Start":"05:56.585 ","End":"05:58.295","Text":"That is our final answer."},{"Start":"05:58.295 ","End":"06:00.960","Text":"Thank you very much for watching."}],"ID":23798},{"Watched":false,"Name":"Reaction Yields - Theory","Duration":"3m 34s","ChapterTopicVideoID":16927,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:04.260","Text":"In previous videos, we use the combustion of"},{"Start":"00:04.260 ","End":"00:07.545","Text":"glucose to explain several portrait concepts."},{"Start":"00:07.545 ","End":"00:09.855","Text":"Here\u0027s the combustion of glucose."},{"Start":"00:09.855 ","End":"00:13.620","Text":"We saw that 1 mole of glucose reacts with"},{"Start":"00:13.620 ","End":"00:17.925","Text":"6 moles of oxygen to give 6 moles of carbon dioxide,"},{"Start":"00:17.925 ","End":"00:19.890","Text":"and 6 moles of water."},{"Start":"00:19.890 ","End":"00:22.710","Text":"We used it for several purposes."},{"Start":"00:22.710 ","End":"00:26.490","Text":"We used it to balance chemical reactions."},{"Start":"00:26.490 ","End":"00:31.875","Text":"We used it to determine the formula of an unknown organic compound."},{"Start":"00:31.875 ","End":"00:35.825","Text":"We used it to talk about stoichiometry."},{"Start":"00:35.825 ","End":"00:40.025","Text":"We used it in our discussion of limiting reactants."},{"Start":"00:40.025 ","End":"00:46.790","Text":"In this video, we\u0027re going to use the same reaction in a discussion of a reaction yield,"},{"Start":"00:46.790 ","End":"00:50.345","Text":"how much of the product do we get in a reaction?"},{"Start":"00:50.345 ","End":"00:53.405","Text":"Let\u0027s begin with the theoretical yield."},{"Start":"00:53.405 ","End":"00:56.330","Text":"The theoretical yield is the amount of the product"},{"Start":"00:56.330 ","End":"00:59.585","Text":"predicted by the stoichiometry of the reaction."},{"Start":"00:59.585 ","End":"01:05.750","Text":"That\u0027s the maximum amount we could possibly obtain if everything goes as it should."},{"Start":"01:05.750 ","End":"01:07.460","Text":"Here\u0027s an example."},{"Start":"01:07.460 ","End":"01:10.280","Text":"Supposing we burn 1 mole of glucose,"},{"Start":"01:10.280 ","End":"01:12.005","Text":"just like in the equation."},{"Start":"01:12.005 ","End":"01:13.790","Text":"According to the equation,"},{"Start":"01:13.790 ","End":"01:17.945","Text":"we should obtain exactly 6 moles of carbon dioxide."},{"Start":"01:17.945 ","End":"01:22.204","Text":"This is called the theoretical yield of carbon dioxide."},{"Start":"01:22.204 ","End":"01:25.895","Text":"It isn\u0027t possible to get more carbon dioxide."},{"Start":"01:25.895 ","End":"01:28.966","Text":"But in fact, often things go wrong,"},{"Start":"01:28.966 ","End":"01:31.475","Text":"and we don\u0027t get the full amount."},{"Start":"01:31.475 ","End":"01:34.820","Text":"We get what we call the actual yield."},{"Start":"01:34.820 ","End":"01:40.070","Text":"The actual yield is the amount of product actually obtained in an experiment."},{"Start":"01:40.070 ","End":"01:44.075","Text":"There can be many reasons why we don\u0027t obtain the full amount."},{"Start":"01:44.075 ","End":"01:45.635","Text":"Perhaps there was a leak,"},{"Start":"01:45.635 ","End":"01:47.900","Text":"perhaps we drop something on the floor."},{"Start":"01:47.900 ","End":"01:52.744","Text":"Perhaps there were other reactions that took away some of our product."},{"Start":"01:52.744 ","End":"01:54.560","Text":"Let\u0027s take an example."},{"Start":"01:54.560 ","End":"01:56.220","Text":"Again, glucose."},{"Start":"01:56.220 ","End":"01:59.705","Text":"Suppose we burn 1 mole of glucose and obtain"},{"Start":"01:59.705 ","End":"02:04.138","Text":"only 5 moles of carbon dioxide because there was a leak,"},{"Start":"02:04.138 ","End":"02:05.855","Text":"and some leaked out."},{"Start":"02:05.855 ","End":"02:08.390","Text":"Recall this, the actual yield,"},{"Start":"02:08.390 ","End":"02:10.880","Text":"the actual amount we obtained."},{"Start":"02:10.880 ","End":"02:14.000","Text":"Now we define the percentage yield,"},{"Start":"02:14.000 ","End":"02:19.085","Text":"the actual percentage that we get out in the reaction."},{"Start":"02:19.085 ","End":"02:20.855","Text":"Here\u0027s the definition."},{"Start":"02:20.855 ","End":"02:24.485","Text":"The percentage yield is the actual yield"},{"Start":"02:24.485 ","End":"02:29.540","Text":"divided by the theoretical yield multiplied by a 100 percent."},{"Start":"02:29.540 ","End":"02:35.150","Text":"If we got all that we should have if the actual yield was equal to the theoretical yield,"},{"Start":"02:35.150 ","End":"02:37.805","Text":"this would give us a 100 percent."},{"Start":"02:37.805 ","End":"02:39.995","Text":"Let\u0027s take the example."},{"Start":"02:39.995 ","End":"02:43.520","Text":"Where the actual yield of carbon dioxide is 5 moles,"},{"Start":"02:43.520 ","End":"02:46.430","Text":"the percentage yield is 5 moles."},{"Start":"02:46.430 ","End":"02:48.020","Text":"That\u0027s the actual yield,"},{"Start":"02:48.020 ","End":"02:50.420","Text":"divided by 6 moles,"},{"Start":"02:50.420 ","End":"02:53.125","Text":"which is the theoretical yield."},{"Start":"02:53.125 ","End":"02:57.686","Text":"And I\u0027ve missed out the moles because they cancel."},{"Start":"02:57.686 ","End":"02:59.225","Text":"Multiplied by a 100 percent."},{"Start":"02:59.225 ","End":"03:04.710","Text":"We work this out, it\u0027s only 83.3 percent."},{"Start":"03:06.700 ","End":"03:13.295","Text":"The percentage yield is significantly smaller than a 100 percent."},{"Start":"03:13.295 ","End":"03:15.395","Text":"This is usually the case."},{"Start":"03:15.395 ","End":"03:19.070","Text":"It\u0027s very rare to get a 100 percent."},{"Start":"03:19.070 ","End":"03:24.065","Text":"In this video, we talked about the theoretical yield,"},{"Start":"03:24.065 ","End":"03:27.485","Text":"the actual yield, and the percentage yield."},{"Start":"03:27.485 ","End":"03:28.850","Text":"In the next video,"},{"Start":"03:28.850 ","End":"03:33.750","Text":"I\u0027ll give a more detailed example of the concept."}],"ID":21151},{"Watched":false,"Name":"Reaction Yields - Example","Duration":"7m 37s","ChapterTopicVideoID":16928,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.530","Text":"In the previous video,"},{"Start":"00:01.530 ","End":"00:04.305","Text":"we discussed the theory of reaction yields."},{"Start":"00:04.305 ","End":"00:07.790","Text":"In this video, I will give an example of how to calculate them."},{"Start":"00:07.790 ","End":"00:09.300","Text":"Here\u0027s our example."},{"Start":"00:09.300 ","End":"00:16.275","Text":"6 grams of solid sulfur reacts with 12 grams of oxygen gas to produce SO_3,"},{"Start":"00:16.275 ","End":"00:19.320","Text":"sulfur trioxide, which is also a gas."},{"Start":"00:19.320 ","End":"00:22.860","Text":"Calculate the theoretical yield of SO_3."},{"Start":"00:22.860 ","End":"00:28.485","Text":"If 12.20 grams is obtained in an experiment,"},{"Start":"00:28.485 ","End":"00:30.810","Text":"calculate the percentage yield."},{"Start":"00:30.810 ","End":"00:32.160","Text":"Before we go on,"},{"Start":"00:32.160 ","End":"00:37.095","Text":"let\u0027s remember that what\u0027s obtained in the experiment is the actual yield."},{"Start":"00:37.095 ","End":"00:40.545","Text":"This number is the actual yield."},{"Start":"00:40.545 ","End":"00:45.710","Text":"The very first step is to write and balance the chemical equation."},{"Start":"00:45.710 ","End":"00:49.955","Text":"S_8, that\u0027s sulfur in its elemental state,"},{"Start":"00:49.955 ","End":"00:54.020","Text":"plus O_2 gives SO_3."},{"Start":"00:54.020 ","End":"00:56.990","Text":"Now, let\u0027s look at the equation."},{"Start":"00:56.990 ","End":"01:02.300","Text":"We see we have 8 sulfurs here and only 1 on the right-hand side,"},{"Start":"01:02.300 ","End":"01:06.490","Text":"so we need to multiply SO_3 by 8."},{"Start":"01:06.490 ","End":"01:09.165","Text":"Let\u0027s look at the oxygen."},{"Start":"01:09.165 ","End":"01:15.470","Text":"We have 8 times 3 oxygens on the right and only 2 on the left,"},{"Start":"01:15.470 ","End":"01:20.790","Text":"so we need 12 times 2 to get to 24."},{"Start":"01:20.790 ","End":"01:24.690","Text":"We have now 24 on the left and 24 on the right,"},{"Start":"01:24.690 ","End":"01:27.390","Text":"so we need to write here 12."},{"Start":"01:27.390 ","End":"01:29.565","Text":"Here, it\u0027s written better."},{"Start":"01:29.565 ","End":"01:33.380","Text":"S_8 solid plus 12 oxygen gas,"},{"Start":"01:33.380 ","End":"01:36.185","Text":"giving us 8 SO_3 gas."},{"Start":"01:36.185 ","End":"01:39.935","Text":"Now, we can write out our scheme for solving the problem."},{"Start":"01:39.935 ","End":"01:42.080","Text":"We\u0027re given the mass of reactants,"},{"Start":"01:42.080 ","End":"01:44.270","Text":"we can calculate the moles of reactants."},{"Start":"01:44.270 ","End":"01:46.265","Text":"That\u0027s step 2."},{"Start":"01:46.265 ","End":"01:53.375","Text":"Then we have to decide whether there\u0027s a limiting reactant, that step 2a."},{"Start":"01:53.375 ","End":"01:58.625","Text":"Then we\u0027d go from the moles of reactants to the moles of products in here,"},{"Start":"01:58.625 ","End":"02:01.595","Text":"counter the limiting reactant if there is one."},{"Start":"02:01.595 ","End":"02:03.290","Text":"That\u0027s step 3."},{"Start":"02:03.290 ","End":"02:07.238","Text":"For the moles of the product to the mass of product, that\u0027s step 4."},{"Start":"02:07.238 ","End":"02:13.505","Text":"The mass of the product given us here will be the theoretical yield."},{"Start":"02:13.505 ","End":"02:16.535","Text":"Then finally, step 5,"},{"Start":"02:16.535 ","End":"02:21.360","Text":"regards to the percentage yield. Let\u0027s do it."},{"Start":"02:22.250 ","End":"02:27.365","Text":"Step 2, we have to calculate the moles of reactants."},{"Start":"02:27.365 ","End":"02:28.565","Text":"We\u0027re given the masses,"},{"Start":"02:28.565 ","End":"02:31.265","Text":"we need to convert them to moles."},{"Start":"02:31.265 ","End":"02:34.040","Text":"We need to know the molar masses."},{"Start":"02:34.040 ","End":"02:38.479","Text":"The molar mass of S_8 is 256.5 grams per mole,"},{"Start":"02:38.479 ","End":"02:43.490","Text":"and the molar mass of O_2 is 32.00 grams per mole."},{"Start":"02:43.490 ","End":"02:46.190","Text":"Now, we can calculate the number of moles."},{"Start":"02:46.190 ","End":"02:52.840","Text":"The number of moles of acid is the mass of acid divided by the molar mass,"},{"Start":"02:52.840 ","End":"02:58.988","Text":"so 6.00 grams divided by 256.5 grams per mole,"},{"Start":"02:58.988 ","End":"03:03.685","Text":"and that gives us 0.02339 moles."},{"Start":"03:03.685 ","End":"03:06.640","Text":"The number of moles of oxygen is 12 grams,"},{"Start":"03:06.640 ","End":"03:08.935","Text":"that\u0027s how much the mass of oxygen,"},{"Start":"03:08.935 ","End":"03:10.630","Text":"divided by the molar mass,"},{"Start":"03:10.630 ","End":"03:13.285","Text":"32.00 grams per mole."},{"Start":"03:13.285 ","End":"03:15.415","Text":"Again, the grams goes,"},{"Start":"03:15.415 ","End":"03:17.365","Text":"grams cancels with grams,"},{"Start":"03:17.365 ","End":"03:21.220","Text":"we\u0027re left with 1 over moles minus 1 which is moles."},{"Start":"03:21.220 ","End":"03:28.360","Text":"So 12 divided by 32 gives us 0.3750 moles."},{"Start":"03:28.360 ","End":"03:34.190","Text":"We have to look and see whether we have a limiting reactant or a limiting reagent,"},{"Start":"03:34.190 ","End":"03:35.750","Text":"some people call it."},{"Start":"03:35.750 ","End":"03:39.710","Text":"Supposing we want to use all the oxygen,"},{"Start":"03:39.710 ","End":"03:43.865","Text":"we want to use 0.3750 moles of oxygen,"},{"Start":"03:43.865 ","End":"03:47.975","Text":"how many moles of sulfur would we require?"},{"Start":"03:47.975 ","End":"03:52.565","Text":"The moles of sulfur is equal to the moles of oxygen"},{"Start":"03:52.565 ","End":"03:59.600","Text":"multiplied by the stoichiometric ratio, or stoichiometric factor."},{"Start":"03:59.600 ","End":"04:06.455","Text":"1 mole of S_8 requires 12 moles of oxygen to react."},{"Start":"04:06.455 ","End":"04:08.300","Text":"Now we can calculate it."},{"Start":"04:08.300 ","End":"04:14.540","Text":"We can substitute the value of the number of moles of oxygen, 0.3750."},{"Start":"04:14.540 ","End":"04:19.615","Text":"Here\u0027s 0.3750 divided by 12,"},{"Start":"04:19.615 ","End":"04:24.830","Text":"and the units are moles of sulfur because moles of oxygen goes with moles of the oxygen,"},{"Start":"04:24.830 ","End":"04:26.870","Text":"we\u0027re left with moles of sulfur."},{"Start":"04:26.870 ","End":"04:31.400","Text":"We have 0.300750 divided by 12 moles of sulfur,"},{"Start":"04:31.400 ","End":"04:37.870","Text":"and if we divide that we get 0.03125 moles of sulfur."},{"Start":"04:37.870 ","End":"04:42.140","Text":"This tells us that in order to use up all the oxygen,"},{"Start":"04:42.140 ","End":"04:46.789","Text":"we require 0.03125 moles of sulfur."},{"Start":"04:46.789 ","End":"04:50.135","Text":"In the step before the second step,"},{"Start":"04:50.135 ","End":"04:51.560","Text":"we saw this,"},{"Start":"04:51.560 ","End":"04:55.790","Text":"in fact we only have 0.02339 moles"},{"Start":"04:55.790 ","End":"05:00.920","Text":"so we don\u0027t have enough sulfur for all the oxygen to react."},{"Start":"05:00.920 ","End":"05:03.590","Text":"Sulfur is the limiting reactant;"},{"Start":"05:03.590 ","End":"05:07.565","Text":"it will all be used up and some oxygen will be left."},{"Start":"05:07.565 ","End":"05:09.695","Text":"Oxygen is in excess."},{"Start":"05:09.695 ","End":"05:13.460","Text":"Step 3 is to calculate the moles of SO_3."},{"Start":"05:13.460 ","End":"05:21.170","Text":"The moles of SO_3 is number of moles of sulfur multiplied by the stoichiometric factor."},{"Start":"05:21.170 ","End":"05:23.195","Text":"For every mole of S_8,"},{"Start":"05:23.195 ","End":"05:25.895","Text":"we get 8 moles of SO_3."},{"Start":"05:25.895 ","End":"05:27.470","Text":"So it\u0027s 8 to 1."},{"Start":"05:27.470 ","End":"05:30.544","Text":"Now, we can substitute the numbers."},{"Start":"05:30.544 ","End":"05:37.460","Text":"The number of moles of S_8 that we have in practice is 0.02339 moles,"},{"Start":"05:37.460 ","End":"05:43.550","Text":"and that\u0027s multiplied by 8, giving us 0.1871."},{"Start":"05:43.550 ","End":"05:45.245","Text":"What about the units?"},{"Start":"05:45.245 ","End":"05:48.425","Text":"Moles of sulfur goes with moles of sulfur,"},{"Start":"05:48.425 ","End":"05:51.200","Text":"moles of sulfur here goes from moles of sulfur,"},{"Start":"05:51.200 ","End":"05:55.330","Text":"so we\u0027re left with just moles of SO_3."},{"Start":"05:55.330 ","End":"06:01.860","Text":"Our answer is 0.1871 moles of SO_3."},{"Start":"06:01.860 ","End":"06:06.750","Text":"We need to go from moles SO_3 to the mass of SO_3."},{"Start":"06:06.750 ","End":"06:09.935","Text":"To do that, we need the molar mass."},{"Start":"06:09.935 ","End":"06:14.700","Text":"The molar mass of SO_3 is 80.07 grams per mole,"},{"Start":"06:14.700 ","End":"06:17.510","Text":"and the mass, remember,"},{"Start":"06:17.510 ","End":"06:20.255","Text":"is the number of moles times the molar mass."},{"Start":"06:20.255 ","End":"06:25.355","Text":"That\u0027s 0.1871 moles as we calculated before,"},{"Start":"06:25.355 ","End":"06:28.760","Text":"times 80.07 grams per mole."},{"Start":"06:28.760 ","End":"06:30.695","Text":"Multiply the 2 numbers,"},{"Start":"06:30.695 ","End":"06:38.700","Text":"0.1871 times 80.07, we get 14.98 grams. Why grams?"},{"Start":"06:38.700 ","End":"06:42.405","Text":"Because moles times moles^minus 1 is just 1,"},{"Start":"06:42.405 ","End":"06:44.625","Text":"and we\u0027re left just with grams."},{"Start":"06:44.625 ","End":"06:48.605","Text":"Here\u0027s our answer, 14.98 grams."},{"Start":"06:48.605 ","End":"06:51.925","Text":"That is the theoretical yield."},{"Start":"06:51.925 ","End":"06:56.965","Text":"All that remains now is to calculate the percentage yield."},{"Start":"06:56.965 ","End":"06:59.840","Text":"The percentage yield is the actual yield,"},{"Start":"06:59.840 ","End":"07:03.545","Text":"that\u0027s what\u0027s given to us in the question, 12.20 grams,"},{"Start":"07:03.545 ","End":"07:08.540","Text":"divided by the theoretical yield, 14.98 grams,"},{"Start":"07:08.540 ","End":"07:11.620","Text":"which we just calculated in the step before,"},{"Start":"07:11.620 ","End":"07:13.770","Text":"times 100 percent,"},{"Start":"07:13.770 ","End":"07:16.935","Text":"and that\u0027s 81.44 percent."},{"Start":"07:16.935 ","End":"07:22.000","Text":"Here\u0027s our percentage yield, 81.44 percent."},{"Start":"07:22.000 ","End":"07:27.230","Text":"In this video, we solved an example based on the reaction yield."},{"Start":"07:27.230 ","End":"07:32.645","Text":"The solution involved many concepts that we\u0027ve learned in previous videos."},{"Start":"07:32.645 ","End":"07:36.780","Text":"It\u0027s really quite a good revision of those concepts."}],"ID":21152},{"Watched":false,"Name":"Exercise 22","Duration":"3m 15s","ChapterTopicVideoID":23399,"CourseChapterTopicPlaylistID":80086,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.475","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.475 ","End":"00:08.640","Text":"In the reaction of 1.9 mole carbon tetrachloride with an excess of hydrofluoric acid,"},{"Start":"00:08.640 ","End":"00:13.155","Text":"1.45 mole of dichlorodifluoromethane is obtained."},{"Start":"00:13.155 ","End":"00:15.105","Text":"What are the theoretical,"},{"Start":"00:15.105 ","End":"00:17.730","Text":"actual and percent yields of this reaction?"},{"Start":"00:17.730 ","End":"00:21.855","Text":"Carbon tetrachloride plus 2 hydrofluoric acid gives us"},{"Start":"00:21.855 ","End":"00:26.370","Text":"dichlorodifluoromethane plus 2 hydrochloric acid."},{"Start":"00:26.370 ","End":"00:29.040","Text":"First, we\u0027re going to calculate our theoretical yield."},{"Start":"00:29.040 ","End":"00:32.355","Text":"Theoretical yield of a reaction is the calculated quantity of"},{"Start":"00:32.355 ","End":"00:36.090","Text":"product expected from given quantities of reactant."},{"Start":"00:36.090 ","End":"00:39.180","Text":"In our case, we\u0027re going to calculate the number of moles of"},{"Start":"00:39.180 ","End":"00:41.840","Text":"dichlorodifluoromethane from the number"},{"Start":"00:41.840 ","End":"00:44.615","Text":"of moles of the carbon tetrachloride which are given."},{"Start":"00:44.615 ","End":"00:48.745","Text":"The moles of dichlorodifluoromethane"},{"Start":"00:48.745 ","End":"00:53.180","Text":"equal the number of moles of carbon tetrachloride times,"},{"Start":"00:53.180 ","End":"00:54.710","Text":"if we look at our equation,"},{"Start":"00:54.710 ","End":"00:57.575","Text":"we can see that for every 1 mole of carbon tetrachloride,"},{"Start":"00:57.575 ","End":"00:59.765","Text":"we get 1 mole of our product."},{"Start":"00:59.765 ","End":"01:01.650","Text":"It\u0027s times 1 mole of"},{"Start":"01:01.650 ","End":"01:08.180","Text":"dichlorodifluoromethane divided by 1 mole of the carbon tetrachloride."},{"Start":"01:08.180 ","End":"01:10.880","Text":"Now the number of moles of carbon tetrachloride are given and equal"},{"Start":"01:10.880 ","End":"01:15.380","Text":"1.95 mole to this equals 1.95"},{"Start":"01:15.380 ","End":"01:23.165","Text":"mole of carbon tetrachloride times 1 mole of dichlorodifluoromethane,"},{"Start":"01:23.165 ","End":"01:28.095","Text":"divided by 1 mole of carbon tetrachloride."},{"Start":"01:28.095 ","End":"01:31.745","Text":"The moles of carbon tetrachloride cancel out and we get"},{"Start":"01:31.745 ","End":"01:37.165","Text":"1.95 moles of dichlorodifluoromethane."},{"Start":"01:37.165 ","End":"01:40.750","Text":"This is our theoretical yield."},{"Start":"01:45.250 ","End":"01:49.475","Text":"1.95 moles is our theoretical yield."},{"Start":"01:49.475 ","End":"01:54.080","Text":"Now, to go on, the actual yield is the quantity of products which is actually"},{"Start":"01:54.080 ","End":"01:58.520","Text":"produced 4 or 5 moles of dichlorodifluoromethane are obtained,"},{"Start":"01:58.520 ","End":"02:01.225","Text":"and therefore, that is our actual yield."},{"Start":"02:01.225 ","End":"02:05.265","Text":"The moles of the dichlorodifluoromethane,"},{"Start":"02:05.265 ","End":"02:08.790","Text":"which are obtained are 1.45 moles,"},{"Start":"02:08.790 ","End":"02:11.350","Text":"and that is our actual yield."},{"Start":"02:15.010 ","End":"02:18.200","Text":"Now we need to calculate our percent yield."},{"Start":"02:18.200 ","End":"02:25.110","Text":"The percent yield equals"},{"Start":"02:25.120 ","End":"02:31.430","Text":"the actual yield divided by"},{"Start":"02:31.430 ","End":"02:41.280","Text":"the theoretical yield times 100 percent."},{"Start":"02:41.280 ","End":"02:51.110","Text":"In our case, our actual yield is 1.45 mole divided by the theoretical yield,"},{"Start":"02:51.110 ","End":"02:58.700","Text":"which is 1.95 mole times 100 percent."},{"Start":"02:58.700 ","End":"03:05.250","Text":"The moles cancel out and we get 74.36 percent."},{"Start":"03:06.800 ","End":"03:11.735","Text":"Our percent yield equals 74.36 percent."},{"Start":"03:11.735 ","End":"03:13.325","Text":"That is our final answer."},{"Start":"03:13.325 ","End":"03:15.810","Text":"Thank you very much for watching."}],"ID":24288}],"Thumbnail":null,"ID":80086},{"Name":"Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Consecutive Reactions","Duration":"6m 15s","ChapterTopicVideoID":16918,"CourseChapterTopicPlaylistID":80087,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.755","Text":"In this video, we will discuss consecutive reactions."},{"Start":"00:04.755 ","End":"00:10.650","Text":"Consecutive reactions are of the general type A, some reactants,"},{"Start":"00:10.650 ","End":"00:14.295","Text":"going to B some products,"},{"Start":"00:14.295 ","End":"00:19.635","Text":"and then B, being the reactants going to C some products."},{"Start":"00:19.635 ","End":"00:26.130","Text":"B is called an intermediate since it is formed in the first reaction, A to B,"},{"Start":"00:26.130 ","End":"00:28.740","Text":"and then reacts in the second reaction,"},{"Start":"00:28.740 ","End":"00:34.515","Text":"B to C. We can add the 2 equations together to get the overall reaction."},{"Start":"00:34.515 ","End":"00:36.240","Text":"Here are 2 equations,"},{"Start":"00:36.240 ","End":"00:42.950","Text":"A to B and B to C. You see that B appears on the right-hand side in the first equation,"},{"Start":"00:42.950 ","End":"00:44.330","Text":"and on the left-hand side,"},{"Start":"00:44.330 ","End":"00:45.800","Text":"in the second equation."},{"Start":"00:45.800 ","End":"00:47.990","Text":"We can cancel them."},{"Start":"00:47.990 ","End":"00:50.585","Text":"Now we can add up the 2 equations,"},{"Start":"00:50.585 ","End":"00:55.085","Text":"A on the left-hand side and C on the right-hand side."},{"Start":"00:55.085 ","End":"00:57.140","Text":"If that isn\u0027t clear,"},{"Start":"00:57.140 ","End":"00:59.795","Text":"we can do it without canceling the B,"},{"Start":"00:59.795 ","End":"01:04.890","Text":"A plus B going to B plus"},{"Start":"01:04.890 ","End":"01:10.220","Text":"C. Now you can see that B appears in both sides of the equation,"},{"Start":"01:10.220 ","End":"01:14.105","Text":"so can be canceled, leading to A,"},{"Start":"01:14.105 ","End":"01:23.335","Text":"that gives us A going to C. Let\u0027s take an example,"},{"Start":"01:23.335 ","End":"01:28.895","Text":"write the overall reaction for the following balanced consecutive reactions."},{"Start":"01:28.895 ","End":"01:34.210","Text":"CH_4 gas, that\u0027s methane gas plus chlorine gas,"},{"Start":"01:34.210 ","End":"01:36.925","Text":"giving us methyl chloride gas,"},{"Start":"01:36.925 ","End":"01:39.310","Text":"and hydrogen chloride gas."},{"Start":"01:39.310 ","End":"01:46.120","Text":"Then CH_3Cl, that\u0027s methyl chloride reacts with chlorine"},{"Start":"01:46.120 ","End":"01:53.495","Text":"again to give us CH_2CL_2 gas plus HCL gas."},{"Start":"01:53.495 ","End":"02:00.625","Text":"We see that CH_3Cl is formed the first one and reacts in the second reaction."},{"Start":"02:00.625 ","End":"02:09.315","Text":"The second part of the question is if 10 grams of CH_4 reacts with excess CO_2,"},{"Start":"02:09.315 ","End":"02:13.060","Text":"how much CH_2CL_2 will be obtained?"},{"Start":"02:13.060 ","End":"02:16.520","Text":"The first step is to get the overall reaction."},{"Start":"02:16.520 ","End":"02:20.210","Text":"We do that by adding the 2 equations together."},{"Start":"02:20.210 ","End":"02:23.285","Text":"Here are 2 equations as we had before."},{"Start":"02:23.285 ","End":"02:25.445","Text":"Now we can add them together."},{"Start":"02:25.445 ","End":"02:30.175","Text":"We see that CH_3Cl appears on the right here,"},{"Start":"02:30.175 ","End":"02:32.325","Text":"and on the left here."},{"Start":"02:32.325 ","End":"02:33.770","Text":"It appears on the right,"},{"Start":"02:33.770 ","End":"02:35.570","Text":"it\u0027s formed in the first reaction,"},{"Start":"02:35.570 ","End":"02:37.715","Text":"and reacts in the second reaction."},{"Start":"02:37.715 ","End":"02:39.800","Text":"We can cancel it out."},{"Start":"02:39.800 ","End":"02:43.580","Text":"CH_3CL is an intermediate."},{"Start":"02:43.580 ","End":"02:47.000","Text":"Let\u0027s add the 2 to get our overall reaction."},{"Start":"02:47.000 ","End":"02:48.890","Text":"Here we have CH_4,"},{"Start":"02:48.890 ","End":"02:51.060","Text":"CH_4, CL_2,"},{"Start":"02:51.060 ","End":"02:54.075","Text":"and another CL_2 gives us 2 CL_2."},{"Start":"02:54.075 ","End":"02:58.620","Text":"CH_2, CL_2 here it appears to the right hand side."},{"Start":"02:58.620 ","End":"03:04.085","Text":"HCL in the first reaction and HCL in the second reaction gives us 2 HCL."},{"Start":"03:04.085 ","End":"03:10.220","Text":"This reaction is balanced because we added up 2 balanced reactions."},{"Start":"03:10.220 ","End":"03:13.010","Text":"This is the overall reaction."},{"Start":"03:13.010 ","End":"03:14.555","Text":"Now we can write,"},{"Start":"03:14.555 ","End":"03:17.690","Text":"the scheme for answering the question."},{"Start":"03:17.690 ","End":"03:22.500","Text":"The mass of methane goes to moles of methane."},{"Start":"03:22.500 ","End":"03:26.090","Text":"That\u0027s Step 2."},{"Start":"03:26.090 ","End":"03:29.600","Text":"Moles of methane to moles of CH_2CL_2,"},{"Start":"03:29.600 ","End":"03:31.450","Text":"that\u0027s the third step."},{"Start":"03:31.450 ","End":"03:37.320","Text":"The final step, moles of CH_2CL_2 to mass of CH_2CL_2,"},{"Start":"03:37.320 ","End":"03:39.225","Text":"that\u0027s the fourth step."},{"Start":"03:39.225 ","End":"03:43.680","Text":"Step 2 is to turn the mass of CH_4,"},{"Start":"03:43.680 ","End":"03:45.495","Text":"to moles of CH_4."},{"Start":"03:45.495 ","End":"03:48.610","Text":"This is something we\u0027ve done several times."},{"Start":"03:48.610 ","End":"03:51.185","Text":"It should be familiar to you by now."},{"Start":"03:51.185 ","End":"03:55.240","Text":"The first thing we need to know is the molar mass of CH_4."},{"Start":"03:55.240 ","End":"03:58.280","Text":"That\u0027s 16.0 grams per mole."},{"Start":"03:58.280 ","End":"04:00.710","Text":"From that, we can get the number of moles."},{"Start":"04:00.710 ","End":"04:02.630","Text":"The number of moles, if you remember,"},{"Start":"04:02.630 ","End":"04:08.665","Text":"n is equal to the mass divided by the molecular weight or the molar mass."},{"Start":"04:08.665 ","End":"04:12.350","Text":"Here we have it, n_CH_4,"},{"Start":"04:12.350 ","End":"04:19.655","Text":"the number of moles is equal to the mass 10.0 Grams divided by the molar mass,"},{"Start":"04:19.655 ","End":"04:22.055","Text":"16.0 grams per mole."},{"Start":"04:22.055 ","End":"04:25.490","Text":"That gives us 0.625 moles."},{"Start":"04:25.490 ","End":"04:29.280","Text":"Steps 3, the number of moles of CH_4,"},{"Start":"04:29.280 ","End":"04:31.810","Text":"to moles of CH_CL_2."},{"Start":"04:31.810 ","End":"04:34.580","Text":"This is the central part of the problem."},{"Start":"04:34.580 ","End":"04:36.845","Text":"Now if we look at the equation we had,"},{"Start":"04:36.845 ","End":"04:40.880","Text":"we see that for every mole of CH_4,"},{"Start":"04:40.880 ","End":"04:44.405","Text":"we get 1 mole of CH_2CL_2."},{"Start":"04:44.405 ","End":"04:51.580","Text":"The number of moles of CH_2CL_2 will be exactly equal to the number of moles of methane."},{"Start":"04:51.580 ","End":"04:56.390","Text":"N_CH_2CL_2 is the same as the number of moles of methane,"},{"Start":"04:56.390 ","End":"05:02.075","Text":"is 0.625 moles, exactly what we calculated before."},{"Start":"05:02.075 ","End":"05:03.755","Text":"Now in Step 4,"},{"Start":"05:03.755 ","End":"05:11.520","Text":"we calculate the mass of CH_2CL_2 from the number of moles of CH_2CL_2."},{"Start":"05:11.520 ","End":"05:15.655","Text":"Once again, we need the molar mass of CH_2CL_2,"},{"Start":"05:15.655 ","End":"05:19.165","Text":"and that\u0027s 84.9 grams per mole."},{"Start":"05:19.165 ","End":"05:21.410","Text":"We can work out the mass."},{"Start":"05:21.410 ","End":"05:26.000","Text":"We know that the mass is equal to the number of"},{"Start":"05:26.000 ","End":"05:33.100","Text":"moles times the molar mass, that\u0027s Mw."},{"Start":"05:33.100 ","End":"05:35.810","Text":"Here we have the number of moles,"},{"Start":"05:35.810 ","End":"05:39.365","Text":"0.625 moles, and the molar mass,"},{"Start":"05:39.365 ","End":"05:41.930","Text":"84.9 grams per mole."},{"Start":"05:41.930 ","End":"05:47.390","Text":"We can multiply the 2 numbers and get 53.1 and as for the units,"},{"Start":"05:47.390 ","End":"05:51.830","Text":"mole cancels with mole minus 1 to give 1."},{"Start":"05:51.830 ","End":"05:56.529","Text":"We have just the grams remaining, grams, grams."},{"Start":"05:56.529 ","End":"05:59.380","Text":"That\u0027s our final answer."},{"Start":"05:59.380 ","End":"06:04.990","Text":"The mass of CH_2CL_2 is 53.1 grams."},{"Start":"06:04.990 ","End":"06:09.050","Text":"In this video we learned about consecutive reactions,"},{"Start":"06:09.050 ","End":"06:10.730","Text":"in the next video,"},{"Start":"06:10.730 ","End":"06:15.150","Text":"we\u0027ll learn about simultaneous reactions."}],"ID":17665},{"Watched":false,"Name":"Exercise 1","Duration":"10m 36s","ChapterTopicVideoID":31685,"CourseChapterTopicPlaylistID":80087,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33924},{"Watched":false,"Name":"Exercise 2","Duration":"8m 20s","ChapterTopicVideoID":20341,"CourseChapterTopicPlaylistID":80087,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.700","Text":"We\u0027re going to solve the following exercise."},{"Start":"00:02.700 ","End":"00:06.945","Text":"How many moles of chlorine must be consumed in the first reaction to produce"},{"Start":"00:06.945 ","End":"00:11.730","Text":"3.15 kilograms of dichlorodifluoromethane in the second."},{"Start":"00:11.730 ","End":"00:14.250","Text":"Assume that all the carbontetrachloride"},{"Start":"00:14.250 ","End":"00:17.265","Text":"produced in the first reaction is consumed in the second,"},{"Start":"00:17.265 ","End":"00:22.035","Text":"methane plus chlorine gives us carbontetrachloride plus hydrogenchloride,"},{"Start":"00:22.035 ","End":"00:23.430","Text":"and this is not balanced."},{"Start":"00:23.430 ","End":"00:26.760","Text":"Carbontetrachloride plus hydrogenfluoride gives us"},{"Start":"00:26.760 ","End":"00:30.512","Text":"dichlorodifluoromethane plus hydrogenchloride,"},{"Start":"00:30.512 ","End":"00:31.845","Text":"and this is not balanced."},{"Start":"00:31.845 ","End":"00:35.610","Text":"The first step is to balance the reactions."},{"Start":"00:35.610 ","End":"00:38.760","Text":"Let\u0027s write the first one down methane plus"},{"Start":"00:38.760 ","End":"00:43.560","Text":"chlorine gives us carbontetrachloride plus hydrogenchloride."},{"Start":"00:43.560 ","End":"00:46.790","Text":"Which are going to divide this up into the reactants and the products,"},{"Start":"00:46.790 ","End":"00:50.520","Text":"and then the reactants we can see that we have 1 carbon,"},{"Start":"00:50.660 ","End":"00:57.254","Text":"2 chlorines right here and 4 hydrogens."},{"Start":"00:57.254 ","End":"01:01.305","Text":"In the products we can see that we also have 1 carbon,"},{"Start":"01:01.305 ","End":"01:05.940","Text":"we have 5 chlorines and we have 1 hydrogen."},{"Start":"01:05.940 ","End":"01:07.820","Text":"First of all if we look at our carbons there are"},{"Start":"01:07.820 ","End":"01:10.190","Text":"balanced and the chlorine is and hydrogens are not."},{"Start":"01:10.190 ","End":"01:12.870","Text":"However, we have free chlorine right here,"},{"Start":"01:12.870 ","End":"01:16.590","Text":"so it\u0027s going to be very easy to balance the chlorine is at the end."},{"Start":"01:16.590 ","End":"01:18.030","Text":"Let\u0027s get to the hydrogens."},{"Start":"01:18.030 ","End":"01:19.860","Text":"In our reactants we have 4,"},{"Start":"01:19.860 ","End":"01:21.885","Text":"in our products we have 1."},{"Start":"01:21.885 ","End":"01:25.620","Text":"We\u0027re going to multiply hydrogen chloride by 4 that way you\u0027re"},{"Start":"01:25.620 ","End":"01:28.995","Text":"going to change the hydrogens to 4 hydrogens,"},{"Start":"01:28.995 ","End":"01:33.075","Text":"and now we have 4 chlorines from our hydrogen chloride,"},{"Start":"01:33.075 ","End":"01:37.245","Text":"and 4 more chlorines from our carbontetrachloride which is 8 chlorine,"},{"Start":"01:37.245 ","End":"01:40.030","Text":"so you\u0027re going to change the chlorine to 8."},{"Start":"01:40.120 ","End":"01:43.475","Text":"Carbons are balanced, the hydrogens are balanced,"},{"Start":"01:43.475 ","End":"01:44.510","Text":"and chlorine are not."},{"Start":"01:44.510 ","End":"01:49.350","Text":"We have 8 chlorine in the product and 2 chlorines in the reactants,"},{"Start":"01:49.350 ","End":"01:52.260","Text":"therefore, we\u0027re going to multiply our chlorine by 4,"},{"Start":"01:52.260 ","End":"01:55.365","Text":"and then we\u0027re going to have 8 chlorines because it\u0027s 2 times 4."},{"Start":"01:55.365 ","End":"01:57.765","Text":"We\u0027re going to change the chlorines here from 2."},{"Start":"01:57.765 ","End":"02:01.290","Text":"Now we can see that we have 8 chlorines because it\u0027s 2 times 4."},{"Start":"02:01.290 ","End":"02:04.385","Text":"We\u0027re going to change a chlorine here from 2-8."},{"Start":"02:04.385 ","End":"02:06.020","Text":"Now we have 1 carbon,"},{"Start":"02:06.020 ","End":"02:08.299","Text":"8 chlorines and 4 hydrogens,"},{"Start":"02:08.299 ","End":"02:10.025","Text":"and the reaction is balanced."},{"Start":"02:10.025 ","End":"02:11.750","Text":"We balanced the first equation."},{"Start":"02:11.750 ","End":"02:14.030","Text":"Now we have to balance the second equation."},{"Start":"02:14.030 ","End":"02:19.220","Text":"2 carbontetrachloride plus hydrogenfluoride gives"},{"Start":"02:19.220 ","End":"02:24.905","Text":"us dichlorodifluoromethane plus hydrogenchloride."},{"Start":"02:24.905 ","End":"02:28.925","Text":"We\u0027re going to divide this into the reactants on the product side,"},{"Start":"02:28.925 ","End":"02:31.985","Text":"in the reactants we have 1 carbon,"},{"Start":"02:31.985 ","End":"02:40.570","Text":"4 chlorines, we have 1 fluorine and 1 hydrogen."},{"Start":"02:40.850 ","End":"02:45.915","Text":"Now in the product side we have 1 carbon you can see here we have"},{"Start":"02:45.915 ","End":"02:51.210","Text":"3 chlorines we can see 1, 2, 3."},{"Start":"02:51.210 ","End":"02:55.155","Text":"2 from the dichlorodifluoromethane and 1 from the hydrogenchloride."},{"Start":"02:55.155 ","End":"03:02.050","Text":"We have 2 fluorine and we have 1 hydrogen."},{"Start":"03:04.700 ","End":"03:08.730","Text":"As we can see the carbons are balanced,"},{"Start":"03:08.730 ","End":"03:10.595","Text":"now we go to the chlorine."},{"Start":"03:10.595 ","End":"03:15.020","Text":"We can see that we have 4 on the reactant side and 3 on the product side,"},{"Start":"03:15.020 ","End":"03:18.200","Text":"so we want to change the chlorine to 4."},{"Start":"03:18.200 ","End":"03:20.480","Text":"As we can see we have two chlorines in"},{"Start":"03:20.480 ","End":"03:24.950","Text":"our first molecule and we have 1 chlorine in a second one."},{"Start":"03:24.950 ","End":"03:27.620","Text":"We\u0027re going to multiply or hydrogen chloride by 2"},{"Start":"03:27.620 ","End":"03:30.235","Text":"that way we\u0027ll get 2 chlorines from the hydrogenchloride,"},{"Start":"03:30.235 ","End":"03:34.350","Text":"2 from the dichlorodifluoromethane and we\u0027ll have 4 in all."},{"Start":"03:34.350 ","End":"03:36.320","Text":"We\u0027re going to change the chlorine number from"},{"Start":"03:36.320 ","End":"03:40.235","Text":"3-4 and our hydrogen is also changed from 1-2."},{"Start":"03:40.235 ","End":"03:43.360","Text":"We\u0027re going to change our hydrogens from 1-2."},{"Start":"03:43.360 ","End":"03:46.805","Text":"Now carbons are balanced our chlorine is balanced,"},{"Start":"03:46.805 ","End":"03:49.265","Text":"and the fluorine and the hydrogen is not."},{"Start":"03:49.265 ","End":"03:51.470","Text":"We have 2 fluorines and 2 hydrogens in"},{"Start":"03:51.470 ","End":"03:55.265","Text":"our product side and 1 fluorine and 1 hydrogen in reactant side."},{"Start":"03:55.265 ","End":"03:58.940","Text":"We can fix this very quickly by multiplying our hydrogen fluoride by"},{"Start":"03:58.940 ","End":"04:04.465","Text":"2 that will change the 4 into 2 and our hydrogen and also to 2."},{"Start":"04:04.465 ","End":"04:07.320","Text":"Hydrogen 2 fluoride 2."},{"Start":"04:07.320 ","End":"04:11.175","Text":"Now our carbons are balanced our chlorine is balanced,"},{"Start":"04:11.175 ","End":"04:13.890","Text":"our fluoride is balanced and our hydrogen is balanced,"},{"Start":"04:13.890 ","End":"04:15.990","Text":"so reaction is balanced."},{"Start":"04:15.990 ","End":"04:19.175","Text":"Now we\u0027re going to go on. You want to find the number of moles"},{"Start":"04:19.175 ","End":"04:21.875","Text":"of chlorine reaction and we\u0027re given"},{"Start":"04:21.875 ","End":"04:27.770","Text":"the mass of the dichlorodifluoromethane which is produced in the second reaction."},{"Start":"04:27.770 ","End":"04:29.780","Text":"It\u0027s hard to find the number of moles of the chlorine,"},{"Start":"04:29.780 ","End":"04:31.340","Text":"we\u0027re going to first find the number of moles of"},{"Start":"04:31.340 ","End":"04:35.210","Text":"the dichlorodifluoromethane and then find the number of moles of the chlorine."},{"Start":"04:35.210 ","End":"04:39.890","Text":"The number of moles n equals m which is the mass divided by the molar mass."},{"Start":"04:39.890 ","End":"04:44.810","Text":"The mass of the dichlorodifluoromethane is given and equals 3.15 kilograms the mass"},{"Start":"04:44.810 ","End":"04:50.130","Text":"of dichlorodifluoromethane equals 3.15 kilograms."},{"Start":"04:50.130 ","End":"04:54.630","Text":"We\u0027re going to calculate the molar mass of dichlorodifluoromethane."},{"Start":"04:54.970 ","End":"05:01.970","Text":"This equals the molar mass of carbon plus 2 times the molar mass of chlorine since we"},{"Start":"05:01.970 ","End":"05:03.770","Text":"have two chlorine atoms plus"},{"Start":"05:03.770 ","End":"05:08.300","Text":"2 times the molar mass of fluorine since we have two fluorine atoms in the molecule."},{"Start":"05:08.300 ","End":"05:15.215","Text":"This equals 12.01 grams per mole plus 2 times 35.45 grams per mole,"},{"Start":"05:15.215 ","End":"05:17.555","Text":"plus 2 times 19 grams per mole,"},{"Start":"05:17.555 ","End":"05:20.855","Text":"which equals 120.91 grams per mole."},{"Start":"05:20.855 ","End":"05:24.650","Text":"Now the molar masses were taken from the periodic table of elements."},{"Start":"05:24.650 ","End":"05:28.100","Text":"Now we\u0027re going to calculate the number of moles of dichlorodifluoromethane."},{"Start":"05:28.100 ","End":"05:32.850","Text":"The number of moles of dichlorodifluoromethane equals the mass of"},{"Start":"05:32.850 ","End":"05:39.305","Text":"dichlorodifluoromethane divided by the molar mass of dichlorodifluoromethane."},{"Start":"05:39.305 ","End":"05:43.140","Text":"This equals 3.15 kilograms"},{"Start":"05:44.380 ","End":"05:50.765","Text":"divided by 120.91 grams per mole,"},{"Start":"05:50.765 ","End":"05:52.910","Text":"which we just calculated."},{"Start":"05:52.910 ","End":"05:58.080","Text":"Now here we have kilograms at our numerator and then denominator we have grams per mole,"},{"Start":"05:58.080 ","End":"06:00.050","Text":"so we\u0027re going to multiply the numerator by"},{"Start":"06:00.050 ","End":"06:03.980","Text":"a conversion factor in every one kilogram we have 1,000 grams,"},{"Start":"06:03.980 ","End":"06:06.500","Text":"so it\u0027s 1000 grams per one kilogram."},{"Start":"06:06.500 ","End":"06:09.290","Text":"Kilograms cancel out and we\u0027re left with grams,"},{"Start":"06:09.290 ","End":"06:13.675","Text":"so this equals 26.05 mole."},{"Start":"06:13.675 ","End":"06:16.520","Text":"Remember that when you divide by a fraction is the same"},{"Start":"06:16.520 ","End":"06:19.490","Text":"as multiplying by the reciprocal of the fraction,"},{"Start":"06:19.490 ","End":"06:23.105","Text":"in the units we have grams divided by grams per mole,"},{"Start":"06:23.105 ","End":"06:27.105","Text":"so this equals grams times mole."},{"Start":"06:27.105 ","End":"06:30.550","Text":"The grams cancel out and we\u0027re left with moles."},{"Start":"06:30.550 ","End":"06:36.210","Text":"We know that the moles of the dichlorodifluoromethane equal 26.05 mole,"},{"Start":"06:36.210 ","End":"06:40.325","Text":"now we want to find and calculate the moles of chlorine in the first reaction"},{"Start":"06:40.325 ","End":"06:42.980","Text":"and we also know that all the carbontetrachloride"},{"Start":"06:42.980 ","End":"06:44.900","Text":"which is produced in the first reaction,"},{"Start":"06:44.900 ","End":"06:46.480","Text":"is consumed in a second."},{"Start":"06:46.480 ","End":"06:48.365","Text":"If we look at the second reaction,"},{"Start":"06:48.365 ","End":"06:54.154","Text":"we can see that the number of moles of dichlorodifluoromethane which are produced,"},{"Start":"06:54.154 ","End":"06:56.030","Text":"equals the number of moles of"},{"Start":"06:56.030 ","End":"07:00.575","Text":"carbontetrachloride which are consumed and if we look at our first reaction,"},{"Start":"07:00.575 ","End":"07:03.515","Text":"you can see the most of the carbontetrachloride which are"},{"Start":"07:03.515 ","End":"07:08.115","Text":"produced is less than 4 times the number of chlorine moles."},{"Start":"07:08.115 ","End":"07:09.620","Text":"If we look at our first reaction,"},{"Start":"07:09.620 ","End":"07:13.260","Text":"we can see that for every 4 moles of chlorine which react,"},{"Start":"07:13.280 ","End":"07:17.650","Text":"we get 1 mole of carbontetrachloride."},{"Start":"07:17.650 ","End":"07:19.250","Text":"Now if we look at our reactions,"},{"Start":"07:19.250 ","End":"07:23.065","Text":"we can see that in the first reaction for every 4 moles of chlorine which react,"},{"Start":"07:23.065 ","End":"07:26.790","Text":"we get 1 carbontetrachloride mole and if we look at"},{"Start":"07:26.790 ","End":"07:31.175","Text":"our second reaction we can see that for every 1 mole of carbontetrachloride which react,"},{"Start":"07:31.175 ","End":"07:34.700","Text":"we get 1 mole of dichlorodifluoromethane meaning that for"},{"Start":"07:34.700 ","End":"07:39.510","Text":"every 1 mole of chlorine which reacts with 1 mole of dichlorodifluoromethane."},{"Start":"07:39.510 ","End":"07:43.440","Text":"The number of moles of chlorine equal the moles of"},{"Start":"07:43.440 ","End":"07:46.830","Text":"the dichlorodifluoromethane times 4 moles of"},{"Start":"07:46.830 ","End":"07:52.395","Text":"chlorine divided by 1 mole of dichlorodifluoromethane."},{"Start":"07:52.395 ","End":"07:55.665","Text":"This equals 26.05 mole of"},{"Start":"07:55.665 ","End":"07:58.590","Text":"dichlorodifluoromethane times 4 moles of"},{"Start":"07:58.590 ","End":"08:03.285","Text":"chlorine divided by 1 mole of dichlorodifluoromethane."},{"Start":"08:03.285 ","End":"08:12.705","Text":"The moles of dichlorodifluoromethane cancel out and we get 104.2 moles of chlorine."},{"Start":"08:12.705 ","End":"08:16.610","Text":"The number of moles of chlorine equal 104.2 mole."},{"Start":"08:16.610 ","End":"08:18.185","Text":"That is our final answer."},{"Start":"08:18.185 ","End":"08:20.580","Text":"Thank you very much for watching."}],"ID":21158},{"Watched":false,"Name":"Simultaneous Reactions","Duration":"5m 31s","ChapterTopicVideoID":16919,"CourseChapterTopicPlaylistID":80087,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"In the previous video,"},{"Start":"00:01.620 ","End":"00:06.855","Text":"we discussed consecutive reactions where 1 reaction occurs after another."},{"Start":"00:06.855 ","End":"00:09.315","Text":"That is of course consecutively."},{"Start":"00:09.315 ","End":"00:12.945","Text":"In this video, we will discuss simultaneous reactions,"},{"Start":"00:12.945 ","End":"00:17.040","Text":"which occur, not surprisingly, simultaneously."},{"Start":"00:17.040 ","End":"00:20.820","Text":"Simultaneous reactions are 2 or more reactions"},{"Start":"00:20.820 ","End":"00:24.210","Text":"that occur at the same time in the same system."},{"Start":"00:24.210 ","End":"00:25.919","Text":"Let\u0027s take an example."},{"Start":"00:25.919 ","End":"00:34.050","Text":"5 grams of MnO_2 and 5 grams of Mn_3O_4 react with excess CO,"},{"Start":"00:34.050 ","End":"00:35.745","Text":"that\u0027s Carbon monoxide,"},{"Start":"00:35.745 ","End":"00:39.915","Text":"to form MnO and CO_2 carbon dioxide."},{"Start":"00:39.915 ","End":"00:43.520","Text":"Calculate the mass of CO_2 formed."},{"Start":"00:43.520 ","End":"00:48.155","Text":"We\u0027ll begin by writing balanced equations for each reaction."},{"Start":"00:48.155 ","End":"00:50.180","Text":"Here\u0027s the first reaction,"},{"Start":"00:50.180 ","End":"00:52.115","Text":"MnO_2 plus CO,"},{"Start":"00:52.115 ","End":"00:54.695","Text":"giving MnO plus CO_2."},{"Start":"00:54.695 ","End":"00:58.120","Text":"Let\u0027s check whether this equation is balanced."},{"Start":"00:58.120 ","End":"01:00.330","Text":"We\u0027ve 1 Mn on the left-hand side,"},{"Start":"01:00.330 ","End":"01:01.920","Text":"and 1 on the right-hand side."},{"Start":"01:01.920 ","End":"01:05.890","Text":"1 Carbon on the left and 1 Carbon on the right."},{"Start":"01:05.890 ","End":"01:09.560","Text":"2 plus 1 Oxygens on the left,"},{"Start":"01:09.560 ","End":"01:12.365","Text":"and 1 plus 2 Oxygens on the right."},{"Start":"01:12.365 ","End":"01:15.500","Text":"We have 3 on each side, 3 to 3."},{"Start":"01:15.500 ","End":"01:18.305","Text":"This equation is balanced."},{"Start":"01:18.305 ","End":"01:20.405","Text":"Let\u0027s take the next equation."},{"Start":"01:20.405 ","End":"01:24.125","Text":"We have Mn_3O_4 plus CO,"},{"Start":"01:24.125 ","End":"01:26.870","Text":"giving MnO plus CO_2."},{"Start":"01:26.870 ","End":"01:30.355","Text":"Let\u0027s look at the Manganese Mn_3."},{"Start":"01:30.355 ","End":"01:32.145","Text":"We have 3 on the left,"},{"Start":"01:32.145 ","End":"01:33.825","Text":"but only 1 on the right,"},{"Start":"01:33.825 ","End":"01:35.820","Text":"so we have to multiply by 3."},{"Start":"01:35.820 ","End":"01:37.290","Text":"Let\u0027s look at the Carbons,"},{"Start":"01:37.290 ","End":"01:40.555","Text":"we\u0027ve 1 on the left and 1 on the right. That\u0027s okay."},{"Start":"01:40.555 ","End":"01:42.700","Text":"Now look at the Oxygens,"},{"Start":"01:42.700 ","End":"01:45.750","Text":"4 plus 1, that\u0027s 5,"},{"Start":"01:45.750 ","End":"01:47.385","Text":"and the right-hand side,"},{"Start":"01:47.385 ","End":"01:49.455","Text":"3 times 1, is 3,"},{"Start":"01:49.455 ","End":"01:53.540","Text":"plus 2 in carbon dioxide giving us a total of 5."},{"Start":"01:53.540 ","End":"01:56.215","Text":"This equation is balanced."},{"Start":"01:56.215 ","End":"01:59.695","Text":"Now we can write the scheme for solving the problem."},{"Start":"01:59.695 ","End":"02:04.885","Text":"We\u0027re given the mass of the reactants of MnO_2 and Mn_3O_4."},{"Start":"02:04.885 ","End":"02:08.260","Text":"We have to calculate the moles of reactants."},{"Start":"02:08.260 ","End":"02:12.075","Text":"Then from that to the moles of Carbon dioxide,"},{"Start":"02:12.075 ","End":"02:14.850","Text":"and from that to the mass of Carbon dioxide."},{"Start":"02:14.850 ","End":"02:19.130","Text":"We\u0027ll call these steps 2, 3 and 4."},{"Start":"02:19.130 ","End":"02:20.750","Text":"Here\u0027s step 2,"},{"Start":"02:20.750 ","End":"02:23.930","Text":"the masses of reactants to most of reactors."},{"Start":"02:23.930 ","End":"02:28.970","Text":"The molar mass of MnO_2 is 86.94 grams per mole."},{"Start":"02:28.970 ","End":"02:32.105","Text":"From that, we can calculate the number of moles."},{"Start":"02:32.105 ","End":"02:37.835","Text":"Now we know that the number of moles is equal to mass divided by the molar mass."},{"Start":"02:37.835 ","End":"02:41.480","Text":"We have the number of moles of MnO_2, is the mass,"},{"Start":"02:41.480 ","End":"02:45.335","Text":"5 grams divide by 86.94 grams per mole,"},{"Start":"02:45.335 ","End":"02:46.960","Text":"which is the molar mass."},{"Start":"02:46.960 ","End":"02:48.240","Text":"Divide the 2 numbers,"},{"Start":"02:48.240 ","End":"02:51.405","Text":"we get 0.05751,"},{"Start":"02:51.405 ","End":"02:56.555","Text":"grams cancels with grams and 1 over molar minus 1 is mole."},{"Start":"02:56.555 ","End":"03:00.455","Text":"Now we have the number of moles of MnO_2."},{"Start":"03:00.455 ","End":"03:04.715","Text":"Let\u0027s calculate the number of moles of Mn_3O_4."},{"Start":"03:04.715 ","End":"03:07.010","Text":"First of all, we need the molar mass."},{"Start":"03:07.010 ","End":"03:12.060","Text":"That\u0027s 228.8 grams per mole."},{"Start":"03:12.060 ","End":"03:15.430","Text":"I forgot to put grams per mole in."},{"Start":"03:16.340 ","End":"03:19.350","Text":"The number of moles is its mass,"},{"Start":"03:19.350 ","End":"03:27.930","Text":"5.00 divided 228.8 grams per mole."},{"Start":"03:27.930 ","End":"03:30.060","Text":"We divide the 2 numbers,"},{"Start":"03:30.060 ","End":"03:33.270","Text":"we get 0.02185,"},{"Start":"03:33.270 ","End":"03:39.835","Text":"and the grams cancels with grams and 1 over molar minus 1 gives us mole."},{"Start":"03:39.835 ","End":"03:46.295","Text":"Now we know the number of moles of Mn_3O_4."},{"Start":"03:46.295 ","End":"03:47.780","Text":"Now step 3,"},{"Start":"03:47.780 ","End":"03:52.010","Text":"we want to go from the moles of reactants to the moles of CO_2."},{"Start":"03:52.010 ","End":"03:53.730","Text":"If we look at the 2 equations,"},{"Start":"03:53.730 ","End":"03:59.920","Text":"we see that 1 mole of CO2 is produced for 1 mole of Manganese Oxide."},{"Start":"03:59.920 ","End":"04:02.580","Text":"Happens for both equations,"},{"Start":"04:02.580 ","End":"04:08.265","Text":"for MnO_2 and for the 1 for Mn_3O_4."},{"Start":"04:08.265 ","End":"04:11.150","Text":"For that, we can write that the number of moles of"},{"Start":"04:11.150 ","End":"04:14.330","Text":"Carbon dioxide is exactly equal to the number of"},{"Start":"04:14.330 ","End":"04:21.580","Text":"moles of MnO_2 plus the number of moles of Mn_3O_4,"},{"Start":"04:21.580 ","End":"04:28.295","Text":"and we can add these 2 numbers and get 0.07936 moles."},{"Start":"04:28.295 ","End":"04:31.595","Text":"That\u0027s the number of moles of Carbon dioxide"},{"Start":"04:31.595 ","End":"04:37.065","Text":"produced from the 1st and from the 2nd reaction."},{"Start":"04:37.065 ","End":"04:40.285","Text":"The final step, step 4,"},{"Start":"04:40.285 ","End":"04:45.455","Text":"is to calculate the mass of Carbon dioxide from the number of moles."},{"Start":"04:45.455 ","End":"04:50.565","Text":"The molar mass of Carbon dioxide is 44.01 grams per mole."},{"Start":"04:50.565 ","End":"04:54.530","Text":"The mass of Carbon dioxide is a number of moles,"},{"Start":"04:54.530 ","End":"04:59.715","Text":"0.07936 times the molar mass,"},{"Start":"04:59.715 ","End":"05:03.155","Text":"44.01 grams per mole."},{"Start":"05:03.155 ","End":"05:05.060","Text":"We multiply these 2 numbers,"},{"Start":"05:05.060 ","End":"05:07.655","Text":"we get 3.493,"},{"Start":"05:07.655 ","End":"05:13.730","Text":"and the units are just grams because moles times moles to minus 1 is 1."},{"Start":"05:13.730 ","End":"05:16.254","Text":"So we\u0027re left just with grams."},{"Start":"05:16.254 ","End":"05:21.285","Text":"Here\u0027s our final answer, 3.493 grams."},{"Start":"05:21.285 ","End":"05:22.520","Text":"So in this video,"},{"Start":"05:22.520 ","End":"05:26.465","Text":"we discussed simultaneous reactions."},{"Start":"05:26.465 ","End":"05:29.615","Text":"We\u0027ve got CO2 from each reaction,"},{"Start":"05:29.615 ","End":"05:31.830","Text":"and we added the 2."}],"ID":17666}],"Thumbnail":null,"ID":80087}]

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