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3.2 Combustion Stoichiometry 63

For combustion at low temperatures, it is common to assume that the molecular
nitrogen in air is inert. Then the general combustion reaction of a hydrocarbon
C a H b with air is

0
ð
C a H b þ a 0:21O 2 þ0:79N 2 Þ ! bCO 2 þcH 2 O+ dN 2 ð3:7Þ
It is tedious to carry the decimals when AirÞ is referred to as one compound in a
ð
chemical reaction formula. Therefore, in most combustion formula, we see the for-
mula of air being defined by normalizing it against 1 mol of O 2 . And Eq. (3.7) becomes

ð
C a H b þ a O 2 þ3:76N 2 Þ ! bCO 2 þcH 2 O+ dN 2 ð3:8Þ
ð
We have to be careful here that O 2 þ 3:76N 2 Þ is not Airð Þ but 4.76 AirÞ. The
ð
coefficient of 3.76 comes from the mole ratio of N 2 to O 2 in a dry air, which is 0.79/
0.21 = 3.76. And the coefficient 4.76 is from 1=0:21 4:76.
Then we can determine a, b, c, and d. Similar to the approach in Example 3.2,
the mass balances of atoms give:
Carbon balance : a ¼ b
Hydrogen balance : b ¼ 2c
Oxygenbalance : 20:21Þa ¼ 2b þ c
ð
Nitrogen balance : 20:79Þa ¼ 2d
ð
One can easily solve the above four equations for the coefficients:

a ¼ðaþb=4Þ=0:21; b ¼ a; c ¼ b=2; and d ¼ 3:76ða þ b=4Þ
Thus the stoichiometry of a general hydrocarbon C a H b burned with air can be
described as

b b b
C a H b þ aþ ð O 2 þ 3:76N 2 Þ ! aCO 2 þ H 2 O+3:76 aþ N 2 ð3:9Þ
4 2 4

b
ð
The stoichiometric air–fuel ratio is also determined as A=FÞ ¼ 4:76 a þ :
s 4
Example 3.3: Stoichiometric combustion of propane
Determine the stoichiometric combustion equation for propane (C 3 H 8 ) mixture with
dry air.
Solution
For propane C 3 H 8 , a ¼ 3 and b ¼ 8, comparing with Eq. (3.9),
We can get

8 8 8
C 3 H 8 þ 3þ ð O 2 þ3:76N 2 Þ ! 3CO 2 þ H 2 O+3:76 3þ N 2
4 2 4