1. Determine the volume of SO2 (at STP) formed from the reaction of 96.7 mol FeS2 and 55.0 L of O2 at 348 K and 1.20 atm. 4 FeS2(s) + 11O2(g)  2Fe2O3(s) + 8SO2(g)

40.0L of SO2 are produced

Explanation:

To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:

Moles O2:

n = PV/RT

n = 1.20atm*55.0L / 0.082atmL/molK*358K

n = 2.25 moles of O2.

Clearly, limiting reactant is O2.

The moles of SO2 produced are:

2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2

Volume SO2:

V = nRT/P

V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm

V = 40.0L of SO2 are produced

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