Commercial zinc weighing 7 g, containing 7% impurities, was dissolved in an excess of hydrochloric acid.

Commercial zinc weighing 7 g, containing 7% impurities, was dissolved in an excess of hydrochloric acid. Calculate: a) the volume of released hydrogen (n.o.), b) the amount of hydrogen substance.

Given:
m tech. (Zn) = 7 g
ω approx. = 7%

To find:
V (H2) -?
n (H2) -?

1) Zn + 2HCl => ZnCl2 + H2;
2) ω (Zn) = 100% – ω approx. = 100% – 7% = 93%;
3) m clean. (Zn) = ω * m tech. / 100% = 93% * 7/100% = 6.51 g;
4) n (Zn) = m pure. / M = 6.51 / 65 = 0.1 mol;
5) n (H2) = n (Zn) = 0.1 mol;
6) V (H2) = n * Vm = 0.1 * 22.4 = 2.2 liters.

Answer: The volume of H2 is 2.2 liters; the amount of substance is 0.1 mol.