Dissolving 10 g of technical zinc in an excess of dilute hydrochloric acid released 3.1 liters of hydrogen

Dissolving 10 g of technical zinc in an excess of dilute hydrochloric acid released 3.1 liters of hydrogen. Determine the mass fraction of impurities in this zinc sample

Given:
m tech. (Zn) = 10 g
V (H2) = 3.1 L

To find:
ω approx. -?

1) Zn + 2HCl => ZnCl2 + H2;
2) n (H2) = V / Vm = 3.1 / 22.4 = 0.14 mol;
3) n (Zn) = n (H2) = 0.14 mol;
4) m clean. (Zn) = n * M = 0.14 * 65 = 9.1 g;
5) ω (Zn) = m pure. * 100% / m tech. = 9.1 * 100% / 10 = 91%;
6) ω approx. = 100% – ω (Zn) = 100% – 91% = 9%.

Answer: The mass fraction of impurities is 9%.