Determine the mass and volume of carbon dioxide obtained from 200 g of limestone containing 75% calcium carbonate.

The limestone roasting reaction is described by the following equation:

CaCO3 = CaO + CO2 ↑;

When decomposing 1 mol of limestone, 1 mol of calcium oxide and 1 mol of carbon monoxide is obtained.

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 = 200 x 0.75 / 100 = 1.5 mol;

Find the volume of 1.5 mol of carbon dioxide.

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

V CO2 = 1.5 x 22.4 = 33.6 liters;

Determine the weight of 1.5 moles of carbon dioxide.

To do this, we multiply the amount of substance by the weight of 1 mole of gas.

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

m CO2 = 1.5 x 44 = 66 grams;