# Homework Help: Moles remaining after a reaction

1. Jun 21, 2010

### MarcMTL

1. The problem statement, all variables and given/known data

Sulfur dioxide reacts with the oxygen in humid air to produce sulfur trioxide.

After the reaction, how many moles are left of: SO2, O2 and how many moles of SO3 were created?

2. Relevant equations

2 SO2 + O2 --> 2 SO3

3. The attempt at a solution
The limiting reactant is SO2, so we know that it will fully react and we'll end up with 0 moles of SO2.

From previous steps, I know that I have 0.1021 moles of SO2, and 0.0857 moles of O2. I also know that after the reaction, I will have 0.03474 moles of O2 left.

I can't, for the life of me, figure out how many moles of SO3 I'll have at the end.
Shouldn't there be some conservation of mass? I keep finding that I'll have 0.1021 moles of SO3 at the end, which is the same amount as the SO2.

Is this correct? I seem to think that the reaction with O2 will have changed the number of moles.

2. Jun 21, 2010

### Ygggdrasil

Conservation of mass is not the same as conservation of moles. 0.1021 moles of SO3 has more mass than 0.1021 moles of SO2

3. Jun 21, 2010

### redpenguin

Since we know the limiting reactant, the SO2, we can go ahead and solve for the number of mols SO3 that is produced.

From the given information, we can map:
2 SO2 --> 2 SO3
*Note how O2 is not needed to solve this problem

(0.1021mol SO2) X (2mol SO3)/(2mol SO2) = 0.1021mols SO3 (FINAL ANSWER)
*Note that two-to-two molar ratio cancels each other out.
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As a side note:
(Alternatively, if the limiting reactant was not given we would have to pluggin both the SO2 and O2 in separate problems, each to solve for SO3. By comparing the solutions we can determine that the reactant that produces the least SO3 will be our limiting reactant)
Just to give you a deeper understanding of how this works.. I will solve the number of mols SO3 from 02. And it's good practice for you!!

02 --> 2 SO3
*Note that there is a 1:2 molar ratio now and we DO NOT need the information from SO2

(.0857mol 02) X (2mol SO3)/(1mol O2) = .1714mol SO3

By comparing the 0.1021mol SO3 created from SO2 and the 0.1714mol SO3 created from O2, we can conclude that SO2 produces the least SO3 and is our limiting reactant. The theoretical yield SO3 is also implied here as being the smallest solution (the 0.1021mols SO3).