A piece of aluminum was dipped into sulfuric acid weighing 49 g.

A piece of aluminum was dipped into sulfuric acid weighing 49 g. How many grams of salt were formed during this. How much hydrogen is released?

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 sulfuric acid. For this purpose, we divide its weight by the weight of 1 mole of the substance.

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

N H2SO4 = 49/98 = 0.5 mol;

The same amount of hydrogen will be released.

Let’s define 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 = 0.5 x 22.4 = 11.2 liters;

In this case, 0.5 / 3 = 0.1666 mol of aluminum chloride will be synthesized.

Let’s find its weight.

M Al2 (SO4) 3 = 27 x 2 + (32 + 16 x 4) x 3 = 342 grams / mol;

m Al2 (SO4) 3 = 342 x 0.1666 = 56.98 grams;