Volume Question

[tabishis]

Got across this question while studying for Chemistry exam:

Determine the volume of SO2 (at STP) formed from the reaction of 96.7 g of FeS2 and 55.0 L of O2 (at 398 K and 1.20 atm). The molar mass of FeS2 is 119.99 g/mol.

4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)
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I am getting 35.8L for the answer which is not in the answer choices. Can anyone solve it(step-by-step) please? Or just guide me on how to correctly solve this problem.

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[Chowdder]

This is a pretty long question with a lot of calculations...

In any case my answer came out of be approximately 33L, not sure if that's one of the answer choices.

Anyways, assuming that the reaction proceed to completion, then you have to determine the limiting reagent.

First determine the molar quantity of each of the reactants.

For FeS2, it's simply dividing the mass by the molar mass
came out to be approximately 0.81 mols

For O2, I got it by the ideal gas law PV=nRT. Rearrange it and you'll get PV/RT=n
my value came out to approximately 2.02 mols

The equations says that for every 4 mols of FeS2 consumed, 11 mols of O2 is consumed.

If you divide 0.81 by 4 and 2.02 by 11 you'll see that O2 is the limiting reagent.

Then it's a matter of determining how many mols of SO2 is being produced. So for 2.02 mols of O2 consumed, 1.47 mols of SO2 is being produced.

Multiply 1.47 by the molar volume of ideal gas (22.4L/mol) and you'll get 32.9 or approximately 33L.

Is that the answer?

 

[sv3]

what is the right answer? that would help us know if we are right in trying to help you.
i got 33L

 

[tabishis]

33 it is. I am not too sure what I was doing wrong and kept getting 35 for answer. I'll work the problem again and see if I get 33. Thanks a bunch, folks.