What volume of hydrogen will be released during the interaction of 2.7 mol of aluminum with hydrochloric acid?

Aluminum enters into an active reaction with hydrochloric acid. In this case, an aluminum chloride salt is formed and elemental hydrogen is released. The interaction is described by the following chemical equation.

Al + 3HCl = AlCl3 + 3/2 H2;

N Al = 2.7 mol;

The chemical amount of released hydrogen will be: 2.7 x 3/2 = 4.05 mol

Let’s find its volume.

To do this, 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 = 4.05 x 22.4 = 90.72 liters;