How much carbon monoxide (II) is required for the complete reduction of 4 g of iron (III) oxide?

Fe2O3 + 3CO = 2Fe + 3CO2.

m (Fe2O3) = 4 g;

M (Fe2O3) = 159.69 g / mol;

n = m / M.

n (Fe2O3) = 4 / 159.69 = 0.025 mol.

According to the stoichiometric equation, we see that CO is 3 times more than Fe2O3, then:

n (CO) = 3 * n (Fe2O3) = 3 * 0.025 = 0.075 mol.

Vm = 22.4 L / mol.

Vm = V / n.

V = Vm * n.

V (CO) = Vm * n (CO) = 22.4 * 0.075 = 1.68l.