Calculate the volume of H2 released during the interaction of 6.5 g of zinc with HCL.

Let’s write the reaction equation:
Zn + 2HCl = ZnCl2 + H2 ↑
Let’s find the amount of zinc substance:
v (Zn) = m (Zn) / M (Zn) = 6.5 / 65 = 0.1 (mol).
According to the reaction equation, 1 mole of H2 is formed per 1 mole of Zn, therefore:
v (H2) = v (Zn) = 0.1 (mol).
Thus, the volume of released hydrogen, measured under normal conditions (n.o.):
V (H2) = v (H2) * Vm = 0.1 * 22.4 = 2.24 (l).
Answer: 2.24 l.