How many liters of hydrogen will be released when 65 g of 20% hydrochloric acid reacts with 15 g of zinc?

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 = 15/65 = 0.2308 mol;

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

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

N HCl = 65 x 0.2 / 36.5 = 0.3562 mol;

Metal is taken in excess

0.3562 mol of acid is reacted with 0.3562 / 2 = 0.1781 mol of metal, while 0.1781 mol of hydrogen is released.

Let’s calculate the volume that it occupies.

To this end, we multiply the amount of substance by the volume of 1 mole of gas (assuming a volume of 22.4 liters).

V H2 = 0.1781 x 22.4 = 3.99 liters;