Calculate the volume of hydrogen obtained by the interaction of zinc and 200 g of hydrochloric acid

univerkov | December 3, 2020 | education

Calculate the volume of hydrogen obtained by the interaction of zinc and 200 g of hydrochloric acid with a mass fraction of HCL equal to 14%.

Zn + 2HCl = ZnCl2 + H2
m (HCl) = m (solution) * w = 0.14 * 200 = 28 g
n (HCl) = m \ M = 28 \ 36.5 = 0.77 mol
n (H2) = 0.77: 2 = 0.385 mol
V (H2) = n * Vm = 0.385 * 22.4 = 8.624 l
Answer: V (H2) = 8.624 l

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