Question

1.     The combustion of methane can be written as: C₃H₈ + 5O₂ == 3CO₂ + 4H₂O. Determine (a) the theoretical combustion air, (b) the excess combustion air at 100% excess rate, and (c) the actual combustion air. Report the amount per lb-mole of C₃H₈ and per lb of C₃H₈.

Answer 1

(a) Since the complete combustion of one molecule of C₃H₈ requires 5 moles of O2

Molecular weight of C3H8 =44gm=44*0.0081 lbs=0.097lbs

Amount of oxygen consumed=5*32=160gm=0.3534lbs

Therefore oxygen required by one lb of C3H8= 0.353/0.097= 3.64 lbs per lb

Or oxygen required per mole=0.353 lbs of oxygen per mole

(b) In excess of 100%, oxygen required=2*Theoretical combustion air

(c) Actual combustion air= Total amount of gas

Therefore actual combustion=Amount of C3H8 +excess air=1+3.64=4.64lbs