What mass of iron can be obtained by reducing iron from iron oxide 3

What mass of iron can be obtained by reducing iron from iron oxide 3 with a mass of 80 g with hydrogen, if the fraction of the product yield is 80% of the theoretically possible?

Given:
m (Fe2O3) = 80 g
η (Fe) = 80%

Find:
m pract. (Fe) -?

1) Fe2O3 + 3H2 => 2Fe + 3H2O;
2) M (Fe2O3) = Mr (Fe2O3) = Ar (Fe) * N (Fe) + Ar (O) * N (O) = 56 * 2 + 16 * 3 = 160 g / mol;
3) n (Fe2O3) = m (Fe2O3) / M (Fe2O3) = 80/160 = 0.5 mol;
4) n theory. (Fe) = n (Fe2O3) * 2 = 0.5 * 2 = 1 mol;
5) n practical (Fe) = η (Fe) * n theory. (Fe) / 100% = 80% * 1/100% = 0.8 mol;
6) M (Fe) = Ar (Fe) = 56 g / mol;
7) m practical. (Fe) = n practical (Fe) * M (Fe) = 0.8 * 56 = 44.8 g.

Answer: The practical weight of Fe is 44.8 g.