Determine the mass and volume of gas that is released during the interaction of 50 grams

Determine the mass and volume of gas that is released during the interaction of 50 grams of calcium carbonate and hydrochloric acid.

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

CaCO3 + 2HCl = CaCl2 + CO2 + H2O;

1 mol of chalk reacts with 2 mol of acid. This releases 1 mole of carbon dioxide.

Let’s calculate the chemical amount of available limestone.

M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;

N CaCO3 = 50/100 = 0.5 mol;

When this amount of chalk dissolves, 0.5 mol of carbon dioxide will be released.

Let’s calculate its weight and volume.

M CO2 = 12 + 16 x 2 = 44 grams / mol;

m CO2 = 44 x 0.5 = 22 grams;

1 liter of ideal gas assumes a volume of 22.4 liters (under normal conditions).

The gas volume will be:

V CO2 = 0.5 x 22.4 = 11.2 liters;