5.9 g of iron containing 5% impurities entered into the reaction with hydrochloric acid.

5.9 g of iron containing 5% impurities entered into the reaction with hydrochloric acid. how much hydrogen was released in this case?

Given:
m tech. (Fe) = 5.9 g
ω approx. = 5%

Find:
V (H2) -?

Solution:
1) Fe + 2HCl => FeCl2 + H2 ↑;
2) ω (Fe) = 100% – ω (Fe) = 100% – 5% = 95%;
3) m clean. (Fe) = ω (Fe) * m tech. (Fe) / 100% = 95% * 5.9 / 100% = 5.61 g;
4) n (Fe) = m pure. (Fe) / M (Fe) = 5.61 / 56 = 0.1 mol;
5) n (H2) = n (Fe) = 0.1 mol;
6) V (H2) = n (H2) * Vm = 0.1 * 22.4 = 2.2 liters.

Answer: The volume of H2 is 2.2 liters.