What volume of hydrogen will be released during the interaction of 54 g of aluminum and hydrochloric acid?

The reaction of interaction of aluminum with hydrochloric acid is described by the following chemical reaction equation.

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

1 mole of aluminum reacts with three moles of hydrochloric acid. This forms 1 mol of aluminum chloride and 1.5 mol of hydrogen.

Let’s find the amount of a substance contained in 54 grams of aluminum.

To do this, we divide the mass of the metal by its molar mass.

N Al = 54/27 = 2 mol;

When this amount of metal dissolves, 2 x 1.5 = 3 mol of hydrogen gas will be released.

Let’s find its volume.

One mole of ideal gas takes up a volume of 22.4 liters under standard normal conditions.

The volume of hydrogen will be equal to:

V H2 = 3 x 22.4 = 67.2 liters;