What volume of hydrogen is formed by the interaction of 650 g of zinc, containing 20% impurities, with hydrochloric acid?

What volume of hydrogen is formed by the interaction of 650 g of zinc, containing 20% impurities, with hydrochloric acid? (reaction equation: zinc + hydrochloric acid → zinc chloride + hydrogen)

Given:
m tech. (Zn) = 650 g
ω approx. = 20%

To find:
V (H2) -?

1) Zn + 2HCl => ZnCl2 + H2 ↑;
2) ω (Zn) = 100% – ω approx. = 100% – 20% = 80%;
3) m (Zn) = ω * m tech. / 100% = 80% * 650/100% = 520 g;
4) n (Zn) = m / M = 520/65 = 8 mol;
5) n (H2) = n (Zn) = 8 mol;
6) V (H2) = n * Vm = 8 * 22.4 = 179.2 liters.

Answer: The volume of H2 is 179.2 liters.