What amount of slaked lime can be obtained from 200 kg of limestone

What amount of slaked lime can be obtained from 200 kg of limestone containing 25% of impurities, if the yield is 90% of the theoretically possible?

The limestone roasting reaction is described by the following equation:

CaCO3 = CaO + CO2 ↑;

The decomposition of 1 mol of limestone synthesizes 1 mol of calcium oxide and 1 mol of carbon monoxide.

Find the molar amount of calcium carbonate. To do this, divide its weight by its molar weight.

M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol; N CaCO3 = 200,000 x 0.75 / 100 = 1,500 mol;

Let us determine the mass of 1,500 mol of calcium oxide.

For this purpose, we multiply the amount of the substance by its molar weight.

Taking into account the reaction yield of 90%, the mass will be:

M CaO = 40 + 16 = 56 grams / mol; m CaO = 1,500 x 56 x 0.9 = 75,600 grams = 75.6 kg;

Let’s calculate the volume of 108 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 = 108 x 22.4 = 2,419.2 liters;