What volume of carbon monoxide is formed when 100 g of barium carbonate containing 10% impurities is calcined?

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

BaCO3 = BaO + CO2;

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

The weight of pure barium carbonate is 100 x 0.9 = 90 grams.

Let’s determine the chemical amount of a substance in 90 grams of barium carbonate.

M BaCO3 = 137 + 12 + 16 x 3 = 197 grams / mol;

N BaCO3 = 90/197 = 0.457 mol;

Let’s calculate the volume of 0.457 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 = 0.457 x 22.4 = 10.24 liters;