Question

A petroleum gas mixture containing 65% propane (C3H8) and 35% propylene (C3H6) is burned with 40% excess air. The flow rate of the gas mixture is 200 mol/h. All of the propylene and 90% of the propane are consumed. No CO is found in the product. Write the chemical reactions. How much air is added to the reactor (molar flow rate)? What is the molar flow rate of the outlet gas?

Answer 1

flow rate of mixture= 200 mol/hr, Propane =0.65*200= 130 moles/hr and Propylene= 70 moles/ hr

The reactions of combustion of propane are C3H8+ 5O2----> 3CO2+ 4H2O

as per the reaction, one mole of C3H8 requires 5 moles of oxygen and produces 3 mole of CO2 and 4 moles of water

Oxygen required = 5 times moles of oxygen =5*130= 650 mole/hr

Combustion of propylene are C3H6+4.5O2----> 3CO2+ 3H2O, moles of oxygen required= 70*4.5=315 moles/hr

Total oxygen required = 650+315= 965 moles/hr Total air required= 950/0.21 ( air contains 79% N2 and 21% O2 by volume0

Total air required= 4524 moles/ hr Air supplied= 40% excess, hence air supplied =1.4*4524=6334 moles/ hr

Nitrgen in air = 6334*0.79=5004 moles/hr Oxygen supplied= 6334-5004=1330 moles/ hr

All the propylene got combusted propylene in the gas mixture = 70 moles/ hr Oxygen consumed= 70*4.5=315 moles/hr CO2 formed =3*70 moles/ hr =210 moles/hr and water formed =3*70 moles/ hr =210 moles/hr

90% of propane got combusted, propane= 130 moles/hr propane combusted =130*0.9= 117 moles/ hr

propane unreacted =130-117=13 moles/hr Oxygen consumed= 117*5=585 moles/ hr CO2 formed =3*117=351 moles/hr and H2O formed= 3*117=351 moles/ hr, oxygen remaining = total oxygen suppled- oxygen used for

combustino of propylener- oxygen used for combustion of propane= 1330-315-585=430 moles/hr

So products

N2 :5004 Moles/hr

O2 : 430 moles/hr

Proylene :13 moles/hr

CO2 : 210 ( combustion of propylene combustion)+ 351= 561 moles/hr

H2O : 210+351=561 moles/ hr

molar flow rate of products ( or outlet gas) : 5004+430+13+561+561=6569 mol/hr