What volume of carbon monoxide 4 will be released with a volume of 150 g of limestone containing 80%

What volume of carbon monoxide 4 will be released with a volume of 150 g of limestone containing 80% pure calcium carbonate.

The decomposition reaction of limestone is described by the following chemical reaction equation:

CaCO3 = CaO + CO2;

When 1 mol of limestone is decomposed, 1 mol of gaseous carbon dioxide is synthesized.

Let’s calculate the available chemical amount of calcium carbonate substance.

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

N CaCO3 = 150 x 0.8 / 100 = 1.2 mol;

During the reaction, the same amount of carbon dioxide and calcium oxide will be synthesized.

Let’s calculate its volume.

To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance.

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

V CO2 = 1.2 x 22.4 = 26.88 liters;