Calculate the mass of calcium oxide and the volume of carbon dioxide formed when calcining 100 g of limestone containing 10% 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 = 100 x 90% = 90 grams;
Determine the amount of substance in 90 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 = 90/100 = 0.9 mol;
The same amount of carbon monoxide and calcium oxide will be obtained.
Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.
Let’s determine the volume of carbon dioxide:
V CO2 = 0.9 x 22.4 = 20.16 liters;
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 0.9 = 50.4 grams;