Calculate the volume of hydrogen obtained by the interaction of 5.4 aluminum with an excess of hydrochloric acid.

Metallic aluminum interacts with hydrochloric acid. At the same time, the salt of aluminum chloride is synthesized and bubbles of hydrogen gas are released. The reaction is described by the following equation.
Al + 3HCl = AlCl3 + 3/2 H2;
Let’s calculate 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 = 5.4 / 27 = 0.2 mol;
The chemical amount of released hydrogen will be: 0.2 x 3/2 = 0.3 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 = 0.3 x 22.4 = 6.72 liters;