What volume of carbon monoxide (IV) will be released during the decomposition of 400 g

What volume of carbon monoxide (IV) will be released during the decomposition of 400 g of limestone containing 7% 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 = 400 x 93% = 372 grams;

Determine the amount of substance in 372 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 = 372/100 = 3.72 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 = 3.72 x 22.4 = 83.328 liters;