When 18.25 g of technical zinc were dissolved in an excess of dilute hydrochloric acid, 5.6 liters of hydrogen were released

When 18.25 g of technical zinc were dissolved in an excess of dilute hydrochloric acid, 5.6 liters of hydrogen were released. Determine the mass fraction of impurities in this zinc sample.

1) Let’s compose the reaction equation:
Zn + 2HCl = ZnCl2 + H2 ↑
2) Calculate the amount of hydrogen and zinc substance:
n (H2) = V (H2) / Vm = 5.6 / 22.4 = 0.25 mol
according to the reaction equation
n (Zn) = n (H2) = 0.25 mol;
3) Determine the mass of pure zinc and the mass fraction of impurities:
m (Zn) = n (Zn) ∙ M (Zn) = 0.25 ∙ 65 = 16.25 g.
m (impurities) = m (technical Zn) – m (Zn) = 18.25 – 16.25 = 2.0 g.
ω (impurities) = (m (impurities) / m (technical Zn)) ∙ 100% = (2.0 / 18.25) ∙ 100 = 10.96%
Answer: 10.96%.