What is the mass of iron obtained from red iron ore Fe2O3 weighing 80 kg, containing 30% impurities

What is the mass of iron obtained from red iron ore Fe2O3 weighing 80 kg, containing 30% impurities, by the aluminothermal method with a mass fraction of iron yield of 95%?

Given:
m (red iron ore) = 80 kg = 80,000 g
ω approx. = 30%
η (Fe) = 95%

To find:
m (Fe) -?

Decision:
1) Fe2O3 + 2Al => 2Fe + Al2O3;
2) ω (Fe2O3) = 100% – ω approx. = 100% – 30% = 70%;
3) m (Fe2O3) = ω (Fe2O3) * m (red iron ore) / 100% = 70% * 80,000 / 100% = 56,000 g;
4) n (Fe2O3) = m (Fe2O3) / M (Fe2O3) = 56000/160 = 350 mol;
5) n theory. (Fe) = n (Fe2O3) * 2 = 350 * 2 = 700 mol;
6) n practical (Fe) = η (Fe) * n theory. (Fe) / 100% = 95% * 700/100% = 665 mol;
7) m (Fe) = n (Fe) * M (Fe) = 665 * 56 = 37240 g.

Answer: The mass of Fe is 37240 g.