Calculate the volume of hydrogen released by the interaction of 6.5 g of zinc with chloride acid.

Zinc is a bivalent metal. Therefore, during the reaction, it interacts with double the amount of hydrogen chloride.

Zn + 2HCl = ZnCl2 + H2;

Let’s determine the molar amount of metallic zinc. To do this, we divide its weight by the mass of 1 mole of metal.

N Zn = 6.5 / 65 = 0.1 mol;

With this amount of zinc, 0.1 x 2 = 0.2 mol of acid will react. This will synthesize 0.1 mol of hydrogen and 0.1 mol of zinc chloride. Let’s calculate the volume of hydrogen. To this end, we multiply the amount of substance by the volume of 1 mole of gas (22.4 liters).

V H2 = 0.1 x 22.4 = 2.24 liters;