How many liters of hydrogen will be released during the interaction 27g. aluminum with hydrochloric acid?

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

2Al + 6HCl = 2AlCl3 + 3H2;

When two moles of aluminum are dissolved, three moles of hydrogen are formed.

Let’s determine the amount of aluminum substance.

M Al = 27 grams / mol;

N Al = 27/27 = 1 mol;

The amount of hydrogen will be:

N H2 = N Al x 3/2 = 1 x 3/2 = 1.5 mol;

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

The volume of hydrogen will be.

V H2 = 1.5 x 22.4 = 33.6 liters;