What volume of hydrogen is formed by the interaction of hydrochloric acid with 540 mg

What volume of hydrogen is formed by the interaction of hydrochloric acid with 540 mg of aluminum containing 4% of impurities? What amount of salt will be obtained in this case?

Aluminum reacts 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 find the chemical amount of aluminum. For this purpose, we divide its weight by the molar weight of the substance.

M Al = 27 grams / mol;

N Al = 0.54 x 0.96 / 27 = 0.0192 mol;

This will synthesize the same amount of salt.

The chemical amount of released hydrogen will be: 0.0192 x 3/2 = 0.0288 mol

Let’s calculate its volume.

To do this, 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 = 0.0288 x 22.4 = 0.64512 liters;