Determine the volume of hydrogen formed by the interaction of sulfuric acid with aluminum weighing 108 g.

Metallic aluminum interacts with sulfuric acid. In this case, aluminum sulfate is synthesized and hydrogen gas is released. The interaction is described by the following chemical equation.

2Al + 3H2SO4 = Al2 (SO4) 3 + 3 H2;

Let’s find 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 = 108/27 = 4 mol;

The chemical amount of released hydrogen will be: 4 x 3/2 = 6 mol

Let’s define 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 = 6 x 22.4 = 134.4 liters;