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Posted: Thu Mar 12, 2020 2:49 pm
by Baoying Li 1B
For 4.17 part c, how do you know the remaining number of moles if 0.045 mol? What is the approach? Why is it not 0.15 mol as half of the O2 and all of the SO2 was being used?

Re: 4.17

Posted: Thu Mar 12, 2020 2:56 pm
by MingdaH 3B
given the initial .030 mols of SO2 and O2, there should be 0.030 moles of SO3 and 0.015 moles of O2 remaining after the reaction goes to completion, since SO2 is the limiting reactant. The final amount of gas is the amount of O2 added with the newly created SO3.

Re: 4.17

Posted: Thu Mar 12, 2020 3:01 pm
by Julie Park 1G
The total number of moles is 0.045mol after you add the 0.030mol of SO3 and 0.015mol of O together. The reason why only 0.015mol of O2 is involved is because of the presence of a limiting reactant. Basically, for every two moles of SO2, only one mol of O2 can react (according to reaction equation). Therefore, even though there is 0.030mol O2 available, only 0.015mol (half) of it will be used.