What volume of hydrogen is formed by the interaction of hydrochloric acid with 540 mg

What volume of hydrogen is formed by the interaction of hydrochloric acid with 540 mg of aluminum containing 40% of impurities? What amount of the substance of salts will be obtained in this case?

Given:
m tech. (Al) = 540 mg = 0.54 g
ω approx. = 40%

Find:
V (H2) -?
n (salt) -?

Solution:
1) 6HCl + 2Al => 2AlCl3 + 3H2 ↑;
2) M (Al) = Mr (Al) = Ar (Al) = 27 g / mol;
3) ω (Al) = 100% – ω approx. = 100% – 40% = 60%;
4) m clean. (Al) = ω (Al) * m tech. (Al) / 100% = 60% * 0.54 / 100% = 0.324 g;
5) n (Al) = m (Al) / M (Al) = 0.324 / 27 = 0.012 mol;
6) n (H2) = n (Al) * 3/2 = 0.012 * 3/2 = 0.018 mol;
7) V (H2) = n (H2) * Vm = 0.018 * 22.4 = 0.4 l;
8) n (AlCl3) = n (Al) = 0.012 mol.

Answer: The volume of H2 is 0.4 l; the amount of AlCl3 substance is 0.012 mol.