130 gr. zinc reacted with hydrochloric acid. What gas and in what volume was released as a result of the reaction?

Let’s write the reaction equation:

Zn + 2HCl = ZnCl2 + H2 ↑

Let’s find the amount of substance Zn:

v (Zn) = m (Zn) / M (Zn) = 130/65 = 2 (mol).

According to the reaction equation, 1 mole of H2 is formed per 1 mole of Zn, therefore:

v (H2) = v (Zn) = 2 (mol).

Thus, the volume of released hydrogen, measured under normal conditions (n.o.):

V (H2) = v (H2) * Vm = 2 * 22.4 = 44.8 (l).

Answer: hydrogen, 44.8 liters.