17 g of a mixture of iron (2) sulfide and pyrite (FES2) was calcined to form 5.04 L

17 g of a mixture of iron (2) sulfide and pyrite (FES2) was calcined to form 5.04 L of gaseous material. Determine the composition of the initial mixture in mass percent.

Given:
m mixture = 17 g
V (gas) = 5.04 l

To find:
ω (FeS) -?
ω (FeS2) -?

Decision:
1) 4FeS + 7O2 => 2Fe2O3 + 4SO2 ↑;
4FeS2 + 11O2 => 2Fe2O3 + 8SO2 ↑;
2) Let m (FeS) = (x) r, then m (FeS2) = (17 – x) r;
3) n (FeS) = m (FeS) / M (FeS) = (x / 88) mol;
4) n1 (SO2) = n (FeS) = (x / 88) mol;
5) n (FeS2) = m (FeS2) / M (FeS2) = ((17 – x) / 120) mol;
6) n2 (SO2) = n (FeS2) * 2 = ((17 – x) / 120) * 2 = ((17 – x) / 60) mol;
7) n total (SO2) = n1 (SO2) + n1 (SO2) = ((x / 88) + ((17 – x) / 60)) mol;
8) n total (SO2) = V (SO2) / Vm = 5.04 / 22.4 = 0.225 mol;
9) 0.225 = (x / 88) + ((17 – x) / 60);
x = 11;
10) ω (FeS) = m (FeS) * 100% / m mixture = 11 * 100% / 17 = 64.71%;
11) ω (FeS2) = 100% – ω (FeS) = 100% – 64.71% = 35.29%.

Answer: The mass fraction of FeS is 64.71%; FeS2 – 35.29%.