Find the volume of hydrogen in the reaction of 0.45 mol of aluminum with dilute seulfate acid taken in excess.

Metallic aluminum interacts with sulfuric acid. In this case, aluminum sulfate is formed and elemental hydrogen is released. The interaction is described by the following chemical equation.

2Al + 3H2SO4 = Al2 (SO4) 3 + 3 H2;

N Al = 0.45 mol;

The chemical amount of released hydrogen will be: 0.45 x 3/2 = 0.675 mol

Let’s calculate its volume.

For this purpose, we multiply the chemical amount of the substance by the volume of 1 mole of gas (filling a space with a volume of 22.4 liters).

V H2 = 0.675 x 22.4 = 15.12 liters;