How much hydrogen will be released when 6.5 zinc reacts with hydrochloric acid?

How much hydrogen will be released when 6.5 zinc reacts with hydrochloric acid? What is the volume of the released hydrogen?

Zinc interacts with hydrochloric acid according to the following scheme:

Zn + 2HCl → ZnCl2 + H2;

To solve the problem, we calculate the molar amount of zinc.

M Zn = 65 grams / mol;

N Zn = 6.5 / 65 = 0.1 mol;

Therefore, the same amount of hydrogen will be obtained, equal to 0.1 mol.

Its volume will be:

V H2 = 22.4 x 0.1 = 2.24 liters;