Calculate how much limestone needs to be calcined at 75% product yield to get 6 kg of quicklime.

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

CaCO3 = CaO + CO2;

One mole of calcium carbonate forms one mole of calcium oxide and carbon monoxide.

Let us determine the amount of the substance in 6,000 grams of calcium oxide.

Its molar mass is:

M CaO = 40 + 16 = 56 grams / mol;

The amount of the substance will be:

N CaO = 6000/56 = 107.14 mol;

The same amount of limestone must be reacted.

Let’s find its mass.

Its molar mass is:

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

Its mass will be:

m CaCO3 = 107.14 x 100 = 10 714 grams;

With a mass fraction of limestone of 75%, its mass will be:

m limestone = m CaCO3 / 0.75 = 10 714 / 0.75 = 14 285 grams = 14.285 kg.;