How many grams of calcium oxide can be obtained by the interaction of 480 grams of limestone containing 5% impurities.

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.

Determine the mass of pure calcium carbonate.

m CaCO3 = 480 x 95% = 456 grams;

Let’s determine the amount of substance in 456 grams of calcium carbonate.

Its molar mass is:

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

The amount of substance will be:

N CaCO3 = 456/100 = 4.56 mol;

The same amount of calcium oxide will be obtained.

The molar mass of calcium oxide is:

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

The mass of calcium oxide will be:

m CaO = 56 x 4.56 = 255.36 grams;