Find the volume of hydrogen that will be released when interacting with hydrochloric acid 15 g of aluminum containing 30% impurities

Let’s write the reaction equation:
6HCl + 2Al = 2AlCl3 + 3H2

Decision:
1) m (pure Al) = 100% – 30% = 70%
2) m (Al) = 15 g * 70% / 100% = 10.5 g
3) n (Al) = 10.5 g / 27 g / mol = 0.39 mol ≈ 0.4 mol
4) We look at the coefficients and make up the proportion
2 mol (Al) – 3 mol (H2)
0.4 mol (Al) – x mol (H2)
n (H2) = 3 * 0.4 mol / 2 = 0.6 mol
5) V (H2) = 0.6 mol * 22.4 L / mol = 13.44 L
Answer: 13.44 L