What is the volume of hydrogen released during the interaction of 150 ml of sulfuric acid with a density of 1.5 g

What is the volume of hydrogen released during the interaction of 150 ml of sulfuric acid with a density of 1.5 g ml and a mass fraction of 35% with aluminum metal.

The reaction of dissolution of aluminum in sulfuric acid is described by the following chemical reaction equation:

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

Let’s calculate the amount of sulfuric acid available.

Its weight is 150 x 1.5 x 0.35 = 78.75 grams;

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

N H2SO4 = 78.75 / 98 = 0.804 mol;

When this amount of acid reacts, 0.804 mol of hydrogen will be released.

Let’s calculate its volume.

1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

V H2 = 0.804 x 22.4 = 18.01 liters;