How much hydrogen H2 will be released during the interaction of aluminum Al weighing 27 g

How much hydrogen H2 will be released during the interaction of aluminum Al weighing 27 g with hydrochloric acid HCl, weighing 87 g?

Aluminum enters into an active reaction 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 calculate the chemical amount of aluminum. To do this, divide its weight by the weight of 1 mole of the substance.

M Al = 27 grams / mol;

N Al = 27/27 = 1 mol;

Let’s calculate the chemical amount of aluminum chloride. For this purpose, we divide its weight by the molar weight of the substance.

M HCl = 1 + 35.5 = 36.5 grams / mol;

N HCl = 87 / 36.5 = 2.38 mol;

2.38 mol of hydrogen chloride will react with 2.38 / 3 = 0.7933 mol of aluminum. This will release 2.38 / 2 = 1.19 mol of hydrogen.

Let’s define 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 = 1.19 x 22.4 = 26.656 liters;