What volume of hydrogen will be released during the interaction of 10.8 g of Al with hydrochloric acid.

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

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

In accordance with the coefficients of the chemical reaction equation, dissolution of 1 mole of metallic aluminum releases 1.5 moles of hydrogen.

Let’s find the amount of the substance contained in 10.8 grams of aluminum. To do this, we divide the mass of the metal by its molar mass.

M Al = 27 grams / mol;

N Al = 10.8 / 27 = 0.4 mol;

The amount of hydrogen released will be 0.4 x 1.5 = 0.6 mol.

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

The volume of hydrogen will be:

V H2 = 0.6 x 22.4 = 13.44 liters;