When calcining limestone 100 g received quicklime weighing 60 g. Determine the mass fraction of impurities in the limestone.

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.

Find 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 = 100/100 = 1 mol;

Let’s determine the mass of 1 mole of calcium oxide.

To do this, we multiply the amount of the substance by its molar weight.

M CaO = 40 + 16 = 56 grams / mol; m CaO = 1 x 56 = 56 grams;

Since a product weighing 60 grams was obtained, the content of impurities in it is 60 – 56 = 4 grams;

The mass fraction of impurities in the sample is 4/100 = 0.04 = 4%;