What is the volume of CO during thermal decomposition of 500 g of limestone containing 10%

What is the volume of CO during thermal decomposition of 500 g of limestone containing 10% of non-carbonate impurities with a product yield of 60%.

The decomposition reaction of calcium carbonate when heated is described by the following chemical equation:
CaCO3 = CaO + CO2 ↑;
Find the chemical amount of calcium carbonate. To do this, we divide its weight by the mass of 1 mole of the substance.
M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;
N CaCO3 = 500 x 0.9 / 100 = 4.5 mol;
The same molar amount of carbon dioxide can theoretically be obtained.
In order to calculate its volume, you need to multiply the amount of substance by the volume of 1 mole of gas (taking a volume of 22.4 liters). Taking into account the output of 60%, the volume of carbon dioxide will be:
V CO2 = 4.5 x 0.6 x 22.4 = 60.48 liters;