When iron was obtained from iron (III) oxide by the method of aluminothermy, 14 g of iron were obtained.

When iron was obtained from iron (III) oxide by the method of aluminothermy, 14 g of iron were obtained. Determine: a) the mass of consumed iron (III), b) the mass of consumed aluminum.

Let’s find the amount of the substance Fe.

n = m: M.

M (Fe) = 56 g / mol.

n = 14 g: 56 g / mol = 0.25 mol.

Let’s find the quantitative ratios of substances.

Fe2O3 + 2Al = Al2O3 + 2Fe.

For 2 mol of Fe, there is 1 mol of Fe2O3.

Substances are in quantitative ratios of 2: 1.

n (Fe2O3) = ½ n (Fe) = 0.25: 2 = 0.125 mol.

Let’s find the mass of Fe2O3.

M (Fe2O3) = 160 g / mol.

m = 160 g / mol × 0.125 mol = 20 g.

n (Al) = n (Fe) = 0.25 mol

M (Al) = 27g / mol

m = 27 g / mol × 0.25 mol = 6.75 g.

Answer: 20 g; 6.75 g.