What is the volume of carbon monoxide (4) formed during the decomposition of calcium carbonate weighing 10.6 g?

The decomposition reaction of calcium carbonate occurs in accordance with the following chemical reaction equation:

CaCO3 = CaO + CO2;

From 1 mole of calcium carbonate, 1 mole of calcium oxide and 1 mole of carbon dioxide are formed.

Let’s determine the chemical amount of a substance in 10.6 grams of calcium carbonate.

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

N CaCO3 = 10.6 / 100 = 0.106 mol;

Let’s calculate the volume of 0.106 mol of carbon dioxide.

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

The volume of carbon dioxide will be equal to:

V CO2 = 0.106 x 22.4 = 2.374 liters;