Question

4FeS2 + 11O2 > 2Fe2O3 + 8SO2

Given 75.0 grams of iron (IV) sulfide, how many grams of sulfur dioxide are produced?

Answer 1

$\nu=m/M;$

$\nu(FeS2) =75/120=0.625 mol.$

0.625 : 4=$\nu(SO2)$ : 8

$\nu(SO2) =0.625*8/4=1.25 mol.$

$m=\nu*M=1.25*64=80 g.$

Answer: 80 g.