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 stoichiometric air.

Answer 1

Basis : 1 mole of natural gas

It contains 0.998 moles of C3H8 and 0.002 moles of C4H10

combustion reaction of C3H8 is written as C3H8+5O2----> 3CO2+ 4H2O

multiplying the equation with 0.998 gives 0.998C3H8+5*0.998O2-----> 3*0.998CO2+ 4*0.998H2O

but air is used instead of oxygen, hence nitrogen in the feed = 0.79* 5*0.998/0.21=18.7719N2 ( 79% N2 and 21% O2 in air)

hence the balanced equation becomes 0.998C3H8+ 4.99O2+ 18.7719N2--->2.994CO2+3.992H2O+ 18.7719N2

for the combustion of C4H10 is balanced and written as C4H10+6.5O2----> 4CO2+5H2O

mutiplying with 0.002 gives 0.002C4H10+6.5*0.002O2-----> 4*0.002CO2+5*0.002H2O

0.002C4H10+0.013O2----> 0.008CO2+0.010H2O

since oxygen is upplied by merans of ait, moles of nitrogen also enter and this amount is (0.013/0.21)*0.79 =0.489N2

0.002C4H10+0.013O2+0.489N2----> 0.008CO2+0.010H2O+0.489N2