Calculate the mass of sulfuric acid that can be obtained in 3 stages from 1 ton of pyrite

Calculate the mass of sulfuric acid that can be obtained in 3 stages from 1 ton of pyrite (mass fraction of impurities 15%), if the losses at each stage are 10%.

Given:
m tech. (FeS2) = 1 t = 1,000,000 g
ω (approx.) = 15%
η = 90%

To find:
m (H2SO4) -?

1) 4FeS2 + 11O2 => 2Fe2O3 + 8SO2;
2SO2 + O2 => 2SO3;
SO3 + H2O => H2SO4;
2) ω (FeS2) = 100% – ω (approx.) = 100% – 15% = 85%;
3) m clean. (FeS2) = ω * m tech. / 100% = 85% * 1,000,000 / 100% = 850,000 g;
4) n (FeS2) = m pure. / M = 850,000 / 120 = 7083.3 mol;
5) n theory. (SO2) = n (FeS2) * 8/4 = 7083.3 * 8/4 = 14166.6 mol;
6) n practical (SO2) = η * n theory. / 100% = 90% * 14166.6 / 100% = 12749.9 mol;
7) n theory. (SO3) = n practical. (SO2) = 12749.9 mol;
8) n practical (SO3) = η * n theory. / 100% = 90% * 12749.9 / 100% = 11474.9 mol;
9) n theory. (H2SO4) = n practical (SO3) = 11474.9 mol;
10) n practical (H2SO4) = η * n theory. / 100% = 90% * 11474.9 / 100% = 10327.4 mol;
11) m practical. (H2SO4) = n practical * M = 10327.4 * 98 = 1012085 g.

Answer: The practical mass of H2SO4 is 1012085 g.