What volume of hydrogen will be released under the action of 32.5 g of zinc on sulfuric acid if the hydrogen yield is 95%?

The reaction between zinc and sulfuric acid is described by the following chemical reaction equation.

Zn + H2SO4 = ZnSO4 + H2;

1 mole of metal reacts with 1 mole of acid. In this case, 1 mol of salt is synthesized and 1 mol of gaseous hydrogen is released.

Let’s calculate the chemical amount of a substance in 32.5 grams of zinc.

M Zn = 65 grams / mol;

N Zn = 32.5 / 65 = 0.5 mol;

The same chemical amount of hydrogen will be obtained.

Taking into account the reaction yield of 95%, its amount will be 0.5 x 0.95 = 0.475 mol;

Let’s calculate its volume.

One liter of ideal gas under normal conditions assumes a volume of 22.4 liters

V H2 = 0.475 x 22.4 = 10.64 liters;