Simple Complete Combustion QuestionWatch
Part a) I think that I need to find the stoichiometric coefficient for O2 and then divide that by 21 and multiply by 100 to find the theoretical air.
So I went about making an equation for the complete combustion of the gas and ended up with 53 as the coefficient and therefore 252kmol but this is way off.
1 kmol of mixture contains
0.4 kmol of CH4
0.24 kmol of C2H4
0.16 kmol of C3H8
and 0.2 kmol of CO2
Write equations for each combustion
CH4 + 2O2 -> CO2 + 2H2O
C2H4 + 3O2 -> 2CO2 + 2H2O
C3H8 + 5O2 -> 3CO2 + 4H2O
So need 0.4 x 2 + 0.24 x 3 + 0.16 x 5 = 2.32 kmol of O2.
Percentage of O2 in air is 21%, so 2.32/0.21 = 11.048 kmol of air needed
For (b) calculate number of moles of CO2, H2O formed and number of moles of oxygen unused. Also calculate the number of moles of N2, then it is a percentage calculation.
For (c) it is dependent on answer to (b).