# What volume of gas will be released when exposed to 10.42 g of limestone (containing 4% impurities)

**What volume of gas will be released when exposed to 10.42 g of limestone (containing 4% impurities) 36.5 g of 24% hydrochloric acid solution.**

The dissolution reaction of limestone in hydrochloric acid is described by the following chemical reaction equation:

CaCO3 + 2HCl = CaCl2 + CO2 + H2O;

When 1 mol of limestone is dissolved in acid, 1 mol of gaseous carbon dioxide is synthesized. This consumes 2 mol of hydrochloric acid.

Let’s calculate the available chemical amount of hydrochloric acid substance.

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

N HCl = 36.5 x 0.24 / 36.5 = 0.24 mol;

Let’s calculate the available chemical amount of limestone.

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

N CaCO3 = 10.42 x 0.96 / 100 = 0.1 mol;

During the reaction, no more than 0.1 mol of carbon dioxide will be synthesized.

Let’s calculate its volume.

To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance.

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

V CO2 = 0.1 x 22.4 = 2.24 liters;