How many grams of sulfuric acid can be obtained from 1 kg of perite containing 20% impurities?

Given:
m (FeS2) = 1 kg = 1000 g
ω approx. = 20%

To find:
m (H2SO4) -?

Decision:
1) 4FeS2 + 11O2 => 2Fe2O3 + 8SO2;
2SO2 + O2 => 2SO3;
SO3 + H2O => H2SO4;
2) ω (FeS2) = 100% – ω approx. = 100% – 20% = 80%;
3) m clean. (FeS2) = ω (FeS2) * m (FeS2) / 100% = 80% * 1000/100% = 800 g;
4) Mr (FeS2) = Ar (Fe) + Ar (S) * 2 = 56 + 32 * 2 = 120 g / mol;
Mr (H2SO4) = Ar (H) * 2 + Ar (S) + Ar (O) * 4 = 1 * 2 + 32 + 16 * 4 = 98 g / mol;
5) n (FeS2) = m pure. (FeS2) / Mr (FeS2) = 800/120 = 6.7 mol;
6) n (H2SO4) = n (SO3) = n (SO2) = n (FeS2) * 2 = 6.7 * 2 = 13.4 mol;
7) m (H2SO4) = n (H2SO4) * Mr (H2SO4) = 13.4 * 98 = 1313.2 g.

Answer: The mass of H2SO4 is 1313.2 g.