What volume of carbon dioxide can be obtained from limestone weighing 359 kg. In which the mass fraction of impurities is 25%.

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 = 359 x 0.75% = 269.25 kg;
Let’s determine the amount of substance in 269.25 kilograms 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 = 269 250/100 = 2 692.5 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 = 2 692.5 x 22.4 = 60 312 liters;