Find the mass of sulphurous gas, which is formed during the roasting of pyrite weighing

Find the mass of sulphurous gas, which is formed during the roasting of pyrite weighing 750 kg, if the mass fraction of the reaction product yield is 75%

The combustion reaction of pyrite is described by the following chemical reaction equation:

4FeS2 + 11O2 = 2Fe2O3 + 8SO2

From one mole of pyrite, two moles of sulfur dioxide are formed.

Let’s find the amount of substance in 750 kilograms of pyrite.

M FeS2 = 56 + 32 x 2 = 120 grams / mol;

N FeS2 = 750,000/120 = 6,250 mol;

Find the mass 6 250 x 2 = 12 500 mol of sulfur dioxide.

M SO2 = 32 + 16 x 2 = 64 grams / mol;

m SO2 = 64 x 12,500 = 800,000 grams = 800 kg.

Taking into account the reaction yield of 75%, the mass will be:

m SO2 fact = 800 x 0.75 = 600 kg;