What volume of hydrogen is formed by the interaction of 560 g of iron containing 10% of impurities

What volume of hydrogen is formed by the interaction of 560 g of iron containing 10% of impurities with sulfuric acid? (reaction equation: iron + sulfuric acid → iron (II) sulfate + hydrogen)

Given:
m tech. (Fe) = 560 g
ω approx. = 10%

To find:
V (H2) -?

1) Fe + H2SO4 => FeSO4 + H2 ↑;
2) ω (Fe) = 100% – ω approx. = 100% – 10% = 90%;
3) m (Fe) = ω * m tech. / 100% = 90% * 560/100% = 504 g;
4) n (Fe) = m / M = 504/56 = 9 mol;
5) n (H2) = n (Fe) = 9 mol;
6) V (H2) = n * Vm = 9 * 22.4 = 201.6 liters.

Answer: The volume of H2 is 201.6 liters.