Determine the volume of carbon monoxide (IV) that can be obtained from limestone

Determine the volume of carbon monoxide (IV) that can be obtained from limestone weighing 0.5 tons (the mass fraction of CaCO3 in limestone is 95%)

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

CaCO3 = CaO + CO2;

From one mole of calcium carbonate, one mole of calcium oxide and carbon monoxide is formed.

Determine the mass of pure calcium carbonate.

m CaCO3 = 500 x 95% = 475 kg;

Let’s determine the amount of substance in 475 kg of calcium carbonate.

Its molar mass is:

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

The amount of substance will be:

N CaCO3 = 475 x 1000/100 = 4 750 mol;

The same amount of carbon monoxide and calcium oxide will be obtained.

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

Let’s determine the volume of carbon dioxide:

V CO2 = 4 750 x 22.4 = 106 400 liters;