How much hydrogen is required for the interaction of 80 g of iron oxide 3?

How much hydrogen is required for the interaction of 80 g of iron oxide 3? What mass of iron and mass of water is formed in this case?

2Fe2O3 + 3H2 = 4Fe + 3H2O
According to the periodic table, we find the molar masses: M (Fe2O3) = 56 * 2 + 16 * 3 = 160 g / mol; M (Fe) = 56 g / mol; M (H2O) = 16 + 1 * 2 = 18 g / mol.
The amount of iron oxide: n = m / M = 80 g / 160 g / mol = 0.5 mol.
According to the reaction, 2 mol of oxide reacts with 3 mol of H2, n (H2) = 0.5 * 3/2 = 0.75 mol. The volume of hydrogen: V = n * Vm = 0.75 mol * 22.4 l / mol = 16.8 l.
n (Fe) = 0.5 * 4/2 = 1 mol. m (Fe) = n * M = 1 mol * 56 g / mol = 56 g.
n (H2O) = 0.5 * 3/2 = 0.75 mol. m (H2O) = 0.75 mol * 18 g / mol = 13.5 g