Calculate the volume of carbon dioxide obtained by firing 300 kg of magnesium carbonate with a mass

Calculate the volume of carbon dioxide obtained by firing 300 kg of magnesium carbonate with a mass fraction of impurities of 6 percent.

The decomposition reaction of magnesium carbonate is described by the following chemical reaction equation:

MgCO3 = MgO + CO2;

One mole of magnesium carbonate forms one mole of magnesium oxide and one mole of carbon dioxide.

Find the mass of pure magnesium carbonate:

m MgCO3 = 300,000 x 0.94 = 282,000 grams;

Let’s find the amount of substance in this mass.

The molar mass of magnesium carbonate is:

M MgCO3 = 24 + 12 + 16 x 3 = 84 grams / mol;

The amount of magnesium carbonate substance is:

N MgCO3 = 282,000/84 = 3 357 mol;

The same number of moles of carbon monoxide will be obtained.

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

The volume of carbon dioxide will be:

V CO2 = 3 357 x 22.4 = 75 197 liters;