How much hydrogen is released when 27 g of aluminum reacts with sulfate acid?

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 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 = 27/27 = 1 mol;

The chemical amount of released hydrogen will be: 1 x 3/2 = 1.5 mol

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 = 1.5 x 22.4 = 33.6 liters;

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