The volume of hydrogen released during the interaction of 2.7 grams of aluminum and hydrochloric acid.

Aluminum enters into an active reaction with hydrochloric acid. In this case, an aluminum chloride salt can be obtained and hydrogen gas bubbles are released. The reaction is described by the following equation.

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

Let’s determine the chemical amount of aluminum. To do this, divide its weight by the weight of 1 mole of the substance.

M Al = 27 grams / mol;

N Al = 2.70 / 27 = 0.1 mol;

The chemical amount of released hydrogen will be: 0.1 x 3/2 = 0.15 mol

Let’s find its volume.

To this end, 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 = 0.15 x 22.4 = 3.36 liters;