How many liters of carbon monoxide (IV) can be obtained by calcining CaCO3 limestone, weighing 80 g, containing 20% impurities?

The limestone roasting reaction is described by the following equation:

CaCO3 = CaO + CO2 ↑;

The decomposition of 1 mole of limestone synthesizes 1 mole of calcium oxide and 1 mole of carbon monoxide.

Let’s calculate the molar amount of calcium carbonate. To do this, divide its weight by its molar weight.

M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol; N CaCO3 = 80 x 0.8 / 100 = 0.64 mol;

Find the volume of 0.64 mol of carbon dioxide.

To this end, we multiply the amount of substance by the volume of 1 mole of gas (filling 22.4 liters).

V CO2 = 0.64 x 22.4 = 14.34 liters;