What volume of carbon dioxide will be released during the decomposition of 50 g

What volume of carbon dioxide will be released during the decomposition of 50 g of calcium carbonate with 10% impurities; the second product is calcium oxide

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.

The weight of pure calcium carbonate is 50 x 0.1 = 45 grams.

Let’s determine the chemical amount of a substance in 45 grams of calcium carbonate.

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

N CaCO3 = 45/100 = 0.45 mol;

Let’s calculate the volume of 0.45 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 = 0.45 x 22.4 = 84.22 liters;