What volume of hydrogen will be released during the interaction of 16.2 g of aluminum with hydroiodic acid.

Metallic aluminum interacts with hydroiodic acid. In this case, aluminum iodide is formed and hydrogen gas is released. The interaction is described by the following chemical equation.

Al + 3HI = AlI3 + 3/2 H2;

Let’s determine 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 = 16.2 / 27 = 0.6 mol;

The chemical amount of released hydrogen will be: 0.6 x 3/2 = 0.9 mol

Let’s find its volume.

To this end, 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 = 0.9 x 22.4 = 20.16 liters;