Question

FeS2 is fed to a fluid bed roaster at a rate of 300 t/h to dead roast FeS2 to produce SO2 and Fe2O3 according to the reaction 2FeS2 + 5.5O2 = Fe2O3 + 4SO2. Calculate the air flow rate required at 25oC and atmospheric pressure in m3 /h

Answer 1

Molar mass of FeS2 =120 g/mol

mass flow rate of FeS2 = 300 t/h = 300*1000 Kg/hr

molar flow rate of FeS2 = 300*1000/120 = 2500 Kmol/hr

Reaction:

2FeSs2 + 5.5O2 -----> Fe2O3 + 4 SO2

From reaction stoichiometry

2 mol FeS2 = 5.5 mol O2

Hence mols of O2 required = (5.5/2)*2500 = 6875 kmol/hr

Now air contains 21 % O2

Hence Kmol/hr of air theoretically required = 6875/0.21 = 32738.1 kmol/hr

Now we will make use of ideal gas law

P = 1 atm

T = 25 oC = 298.15 K

n = 32738.1 Kmol/hr = 32738.1*1000 mol/hr

R = 0.0821 L atm K-1 mol-1

By ideal gas law

P V = n R T

1*V = 32738.1*1000 *0.0821*298.15

V = 801366860 liter/hr = 801366.9 m3/hr