What mass of magnesium will be reduced by hydrogen with a volume of 44.8 liters taken from magnesium oxide (II)?

According to the condition of the problem, we compose the equation of the recovery process:
MgO + H2 = Mg + H2O – OBP, magnesium is formed;

Calculations by formulas:
M (H2) = 2 g / mol; M (Mg) = 24.3 g / mol.

Proportion:
1 mol of gas at normal level – 22.4 liters;

X mol (H2) -44.8 L from here, X mol (H2) = 1 * 44.8 / 22.4 = 2 mol;

Y (Mg) = 2 mol since the amount of substances is 1 mol.

Find the mass of the metal:
m (Mg) = Y * M = 2 * 24.3 = 48.6 g

Answer: magnesium was obtained with a mass of 48.6 g