Calculate the volume of hydrogen that is formed by the reaction of 13 g of zinc with 219 g of 5% hydrochloric acid.

Metallic zinc “dissolves” in hydrochloric acid, the reaction is described by the following chemical equation:

Zn + 2HCl = ZnCl2 + H2;

Let’s find the chemical amount of zinc. For this purpose, we divide its weight by the molar weight of elemental zinc.

M Zn = 65 grams / mol;

N Zn = 13/65 = 0.2 mol;

Let’s determine the molar amount of hydrogen chloride. To do this, we divide its mass by the molar mass of this elemental acid.

M HCl = 1 + 35.5 = 36.5 grams / mol;

N HCl = 219 x 0.05 / 36.5 = 0.3 mol;

The acid is taken in excess.

0.3 mol of acid is reacted with 0.15 mol of metal, while 0.15 mol of hydrogen is released.

Let’s calculate its volume.

For this purpose, we multiply the amount of substance by the volume of 1 mole of gas (filling the space with 22.4 liters).

V H2 = 0.15 x 22.4 = 3.36 liters;