Question

For a natural gas composed of: C3H8 in volume fraction 0.998 C4H10 in volume fraction 0.002, Please write a balanced chemical reaction for the combustion of this natural gas in 10 mol% sub-stoichiometric air (oxygen and nitrogen).

Answer 1

Solution.

The composition of air is approximated as: oxygen - 21%; nitrogen - 79%. Sub-stoichiometric is fuel-rich air-to-fuel ratio.

A balanced chemical reaction for the combustion of C₃H₈ in pure oxygen is

C₃H₈ + 5 O₂ = 3 CO₂ + 4 H₂O

A balanced chemical reaction for the combustion of C4H10 in pure oxygen is

C₄H₁₀ + 13/2 O₂ = 4 CO₂ + 5 H₂O

A balanced chemical reaction for the combustion of C₃H₈ in the air is (the coefficient before the nitrogen found using a factor 0.79/0.21)

C₃H₈ + 5 O₂ + 18.81 N₂= 3 CO₂ + 4 H₂O + 18.81 N₂

A balanced chemical reaction for the combustion of C4H10 in the air is

C₄H₁₀ + 13/2 O₂ + 24.45 N₂= 4 CO₂ + 5 H₂O+24.45 N₂

Assuming that we have 1 mol of the fuel, the coefficients are:

C₃H₈ + 5 O₂ + 18.81 N₂= 3 CO₂ + 4 H₂O + 18.81 N₂ | x0.998

C₄H₁₀ + 13/2 O₂ + 24.45 N₂= 4 CO₂ + 5 H₂O+24.45 N₂   |   x0.002

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0.998C₃H₈ + 0.002C₄H₁₀ + 5.003O₂ +18.821N₂=3.002CO₂ +4.002H₂O + 18.821N₂

10 mol% sub-stoichiometric air means that the amount of air is 90% from the stoichiometric, therefore,

0.998C₃H₈ + 0.002C₄H₁₀ + 4.503O₂ +16.939N₂=xCO₂ +yCO+zH₂O + 16.939N₂;

x+y = 0.998*3+0.002*4;

2x+y+z =4.503*2

2z = 0.998*8+0.002*10

Solution:

x = 2.002

y = 1.000

z = 4.002

The final balanced equation is

0.998C₃H₈ + 0.002C₄H₁₀ + 4.503O₂ +16.939N₂=2.002CO₂ +1.000CO+4.002H₂O + 16.939N_2.