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Baoying Li 1B
Posts: 113
Joined: Sat Aug 17, 2019 12:18 am


Postby Baoying Li 1B » Thu Mar 12, 2020 2:49 pm

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?

MingdaH 3B
Posts: 133
Joined: Thu Jul 11, 2019 12:17 am

Re: 4.17

Postby MingdaH 3B » Thu Mar 12, 2020 2:56 pm

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.

Julie Park 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

Re: 4.17

Postby Julie Park 1G » Thu Mar 12, 2020 3:01 pm

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.

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