Calculate the volume of carbon monoxide (4) and the mass of calcium oxide that can be obtained

univerkov | February 5, 2021 | education

Calculate the volume of carbon monoxide (4) and the mass of calcium oxide that can be obtained with the complete decomposition of 500 g of limestone (calcium carbonate), which contains 10% of sand 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 = 500 x 90% = 450 grams;

Let’s determine the amount of substance in 450 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 = 450/100 = 4.5 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.

Determine the volume of carbon dioxide:

V CO2 = 4.5 x 22.4 = 1 444.8 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 4.5 = 252 grams;