How many liters of hydrogen will be released when an excess of aluminum interacts with 600 g of a 9.8% sulfuric acid solution?

Metallic aluminum interacts with sulfuric acid. In this case, aluminum sulfate is formed and elemental hydrogen 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 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 = 600 x 0.098 / 98 = 0.6 mol;

The same amount of hydrogen will be released.

Let’s calculate 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 = 0.6 x 22.4 = 13.44 liters;