Determine the volume of carbon dioxide that is formed by the interaction of calcium carbonate

Determine the volume of carbon dioxide that is formed by the interaction of calcium carbonate and hydrochloric acid weighing 7.3 grams.

The reaction of interaction of calcium carbonate with hydrochloric acid is described by the following chemical equation:

CaCO3 + 2HCl = CaCl2 + CO2 + H2O;

1 carbonate molecule interacts with 2 acid molecules. This releases 1 molecule of carbon dioxide.

Let’s calculate the chemical amount of a substance in 7.3 grams of acid.

M HCl = 1 + 35.5 = 36.5 grams / mol;

N HCl = 7.3 / 36.5 = 0.2 mol;

In this case, 0.2 / 2 = 0.1 mol of carbon dioxide will be released.

1 mole of ideal gas normally takes on a volume of 22.40 liters.

V CO2 = 0.1 x 22.40 = 2.24 liters;