# During the interactions of zinc and hydrochloric acid, hydrogen was released with a volume

**During the interactions of zinc and hydrochloric acid, hydrogen was released with a volume of 2.8 liters. How much zinc and hydrochloric acid were taken for the reactions?**

Zinc metal dissolves in hydrochloric acid with the evolution of hydrogen gas and the formation of zinc chloride. The reaction is described by the following equation:

Zn + 2HCl = ZnCl2 + H2;

1 mole of metal reacts with 2 moles of acid. This synthesizes 1 mol of salt and 1 mol of hydrogen.

Let’s calculate the released amount of hydrogen. To do this, we divide the volume of synthesized hydrogen by the volume of 1 mole of gas, which is 22.4 liters.

V H2 = 2.8 liters; N H2 = 2.8 / 22.4 = 0.125 mol;

To synthesize such an amount of hydrogen, it is necessary to take the same amount of zinc. Let’s calculate its weight.

M Zn = 65 grams / mol; m Zn = 0.125 x 65 = 8.125 grams;

To synthesize such an amount of hydrogen, it is necessary to take twice the amount of hydrochloric acid. Let’s calculate its weight.

M HCl = 1 + 35.5 = 36.5 grams / mol; N HCl = 0.125 x 2 x 36.5 = 9.125 mol;