Question

18. One mole of CH₄ is burned completely with 20 moles of air.

What is the stack (flue) gas composition?

(Assume air is 21 mole% O₂, 79% N₂)

Answer 1

CH4 + 2O2 -> CO2 + 2H2O this is the balanced equation of the combustion

1 mol of M -> 20 mol of air

20 mol of air is 79% N2 and 21% O2 so

Mol N2 = 20*0.79 = 15.8 mol N2

Mol O2 = 20*.21 = 4.2 mol O2

lets do a balance:

Species involved:

CH4

inlet: 1 mol,

change: 1 mol reacted

out = 0 mol (all reacted)

O2

inlet: 4.2 mol,

change: 2 mol reacted

out = 2.2 mol are left

N2

inlet: 15.8

change: no change, no reaction of N2

out = 15.8 in and goes out

H2O

inlet: 0 mol,

change: 2 moles reacted for every 1 mol of Methane

out = 2 mol

CO2

inlet: 0 mol,

change: 1 mol formed per 1 mol reacted

out = 1 mol

OUTLET

Total moles: CH4+ O2 + N2 + H2O + CO2 = 0 +2.2 + 15.8 + 2+ 1 = 21

Compositions are just

X = moles of species/ total mole

XCH4 = 0/21 = 0

XCO2 = 1/21 = 0.04

XH2O = 2/21 = 0.095

XO2 = 2.2/21 =0.10

XN2 = 15.8/21 = 0.75

It makes sense that N2 is a lot because it didn't reacted and is present in 80% in air!