# 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;

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