How much hydrogen will be released when 108 g of aluminum reacts with an excess of hydrochloric acid?

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

Al + 3HCl = AlCl3 + 1.5 H2;

When 1 mol of aluminum is dissolved, 1.5 mol of hydrogen is released. This requires 3 mol of hydrochloric acid.

Let’s determine the amount of substance contained in 108 grams of aluminum.

M Al = 27 grams / mol;

N Al = 108/27 = 4 mol;

Reaction with 4 moles of aluminum will produce 6 moles of hydrogen.

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

Let’s determine the volume of hydrogen.

V H2 = 6 x 22.4 = 134.4 liters;