A mixture of aluminum and aluminum oxide weighing 11.25 grams was acted upon by an excess of chloride acid

A mixture of aluminum and aluminum oxide weighing 11.25 grams was acted upon by an excess of chloride acid. At the same time, gas was released with a volume of 10.08. Mass fraction of aluminum in the mixture.

Let’s find the amount of substance H2.

n = V: Vn.

n = 10.08 L: 22.4 L / mol = 0.45 mol.

2Al + 6HCl = 2AlCl3 + 3H2.

There is 2 mol of Al for 3 mol of H2. Substances are in quantitative ratios 3: 2 = 1: 1.5. The amount of Al substance is 1.5 times less than that of H2.

n (Al) = 3 / 2n (H2) = 0.45: 1.5 = 0.3 mol.

Let us find the mass of Al.

m = n × M.

M (Al) = 27 g / mol.

m = 27 g / mol × 0.3 mol = 8.1 g.

Let’s find the mass fraction of aluminum in the mixture.

w = (8.1 g: 11.25 g) × 100% = 72%.

Answer: 72%.