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

September 30, 2021 | education

| **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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