Question

Determine the concentrations (in mole fraction) of all species in the exhaust from the complete combustion of propane (C3H8). Assume 50% excess air (79 mole% N2, 21 mole% O2) is the source for the oxygen.

Answer 1

Balanced chemical reaction of complete combustion of propane

C3H8 + 5O2 +18.8 N2 = 3CO2 + 4 H2O + 18.8 N2

21 mol O2 contains = 79 mol of N2

1 mol O2 contains = 79/21 = 3.76 mol of N2

5 mol of O2 contains = 5*3.76 = 18.8 mol of N2

Now 50% excess air is supplied

C3H8 + 7.5O2 +28.2 N2 = 3CO2 + 4 H2O

Actual N2 supplied = 18.8 x 1.5 = 28.2 mol

Actual O2 supplied = 1.5 x 5 = 7.5 mol

Limiting reactant = C3H8

1 mol C3H8 produces = 3 mol of CO2

1 mol C3H8 produces = 4 mol of H2O

1 mol C3H8 consumed = 5 mol of O2

N2 is an inert gas so it would not be consumed.

In the exhaust stream

Mol of C3H8 = 0 mol

Mol of CO2 = 3 mol

Mol of H2O = 4 mol

Mol of O2 = 7.5 - 5 = 1.5 mol

Mol of N2 = 28.2 mol

Total Moles = 36.7 mol

Mol fraction of CO2 = 3/36.7 = 0.0817 = 8.18%

Mol fraction of H2O = 4/36.7 = 0.1089 = 10.90%

Mol fraction of O2 = 1.5/36.7 = 0.0408 = 4.08%

Mol fraction of N2 = 28.2/36.7 = 0.7683 = 76.84%