# Determine the amount of carbon monoxide substance formed during the decomposition of limestone

Determine the amount of carbon monoxide substance formed during the decomposition of limestone weighing 512.8 grams, the mass fraction of impurities in limestone is 15%

The limestone roasting reaction is described by the following equation:

CaCO3 = CaO + CO2 ↑;

When decomposing 1 mol of limestone, 1 mol of calcium oxide and 1 mol of carbon monoxide is obtained.

Let’s calculate the molar amount of calcium carbonate. For this purpose, we divide its weight by its molar weight.

M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol; N CaCO3 = 512.5 x 0.85 / 100 = 4.1 mol;

Let’s calculate the volume of 4.1 mol of carbon dioxide.

To do this, multiply the amount of substance by the volume of 1 mole of gas (which is 22.4 liters).

V CO2 = 4.1 x 22.4 = 91.84 liters;

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