How much quicklime can be obtained from 200 kg of limestone containing 20% of impurities

How much quicklime can be obtained from 200 kg of limestone containing 20% of impurities, if the product yield is 85%.

The decomposition reaction of limestone (calcium carbonate) is described by the following equation:

CaCO3 = CaO + CO2;

In the course of the reaction, 1 molecule of calcium oxide and 1 molecule of carbon dioxide are synthesized from 1 molecule of calcium carbonate.

Let’s calculate the amount of substance contained in 200 kg of chalk.

To do this, divide the available weight of chalk by the weight of 1 mole of chalk.

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

N CaCO3 = 200,000 x 0.8 / 100 = 1,600 mol;

Taking into account the yield of 85% during the reaction, 1600 x 0.85 = 1360 mol of calcium oxide can be synthesized. Let’s calculate its weight.

To do this, we multiply the amount of the substance by the weight of 1 mole of the substance.

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

m CaO = 1360 x 56 = 76 160 grams = 76.16 kg.;