The volume of hydrogen released during the interaction of aluminum weighing 2.7 g with a solution

The volume of hydrogen released during the interaction of aluminum weighing 2.7 g with a solution of sulfuric acid weighing 392 g with a mass fraction of acid of 10%, if the hydrogen yield is 90%

Metallic aluminum interacts with sulfuric acid. This results in the formation of aluminum sulfate and the evolution of hydrogen gas. The interaction is described by the following chemical equation.

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

Let’s find the chemical amount of sulfuric acid. To do this, we divide its weight by the molar weight of the substance.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;

N H2SO4 = 392 x 0.1 / 98 = 0.4 mol;

Let’s calculate the chemical amount of aluminum. For this purpose, we divide its weight by the weight of 1 mole of the substance.

M Al = 27 grams / mol;

N Al = 2.7 / 27 = 0.1 mol;

With 0.1 mol of aluminum, 0.1 x 3/2 = 0.15 mol of sulfuric acid will react.

Taking into account the reaction yield of 90%, 0.15 x 0.9 = 0.135 mol of hydrogen will be released.

Let’s calculate its volume.

To do this, 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 = 0.135 x 22.4 = 3.024 liters;