Find the volume of gas obtained by the interaction of 130 g of zinc with hydrochloric acid. Gas yield 75%.

univerkov | January 17, 2021 | education

Given:
m (Zn) = 130 g
η (H2) = 75% = 0.75
Find: V prak (H2)
Decision:
Zn + 2HCl = ZnCl2 + H2
n (Zn) = m / M = 130 g / 65 g / mol = 2 mol
n (Zn): n (H2) = 1: 1
n (H2) = 2 mol
V theor. (H2) = n * Vm = 2 mol * 22.4 l / mol = 44.8 l
V practical (H2) = η * V theory = 0.75 * 44.8 l = 33.6 l
Answer: 33.6 L

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