During thermal decomposition of a piece of limestone weighing 150 gr. with a mass fraction of calcium carbonate 80%

During thermal decomposition of a piece of limestone weighing 150 gr. with a mass fraction of calcium carbonate 80%, a gas with a volume of 25 liters was released. The reaction yield was …?

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

CaCO3 = CaO + CO2 ↑;

Let’s calculate the chemical amount of calcium carbonate (limestone). To do this, divide its weight by the mass of 1 mole of the substance.

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

N CaCO3 = 150/100 = 1.5 mol;

The same molar amount of carbon dioxide can theoretically be obtained.

To find its volume, you need to multiply the amount of substance by the volume of 1 mole of gas (which is 22.4 liters).

V CO2 = 1.5 x 22.4 = 33.6 liters;

The volume fraction of the reaction yield will be:

K = 25 / 33.6 = 0.744 = 74.4%;