[natalie_gs]

A gas mixture has the following composition ; = 24.0% ;
= 16% and 

(a) How many kmol of air are theoretically required to burn 1 kmol of the gas
mixture?

(b) If the mixture is burned with 20% excess air and complete combustion takes
place, what is the full percentage composition of the
ue gas?

Really need help, no idea were to start with this,

any help appreciated

Thank you.

[Nicolas88]

first you need to know how many moles of oxygen every reactant needs:

CH₄+2O₂ CO₂+2H₂O
C₂H₄+3O₂ 2CO₂+2H₂O
C₃H₈+5O₂ 3CO₂+4H₂O

now we combine them in one equation but divide each of the gases to its ratio:
0.4CH₄+0.24C₂H₄+0.16C₃H₈+(0.4*2)O₂+(0.24*3)O₂+(0.16*5)O₂+0.2CO₂ 0.4CO₂+(2*0.24)CO₂ +(3*0.16)CO₂+(0.4*2)H₂O+(0.24*2)H₂O+(0.16*4)H₂O

now you need to add the mole ratios of oxygen together then you will get how many moles you need to burn 1 mole of the gas, and as you know there is 20% of oxygen in air so you can calculate the moles of air needed
check the answer cause i might be wrong.