Composition of flue gases by volume on a wet basis and dry basis

[dbomb1203]

Homework Statement

A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25ºC, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90ºC

(e) Determine the composition of the flue gases by volume (assuming the inlet air is dry) :
(i) on a wet basis
(ii) on a dry basis.

I'm aware that there is a similar question already posted from 6+ years ago. I have read that particular forum and seem to still struggle.

Relevant Equations

Balanced reactions
Butane C4H10 + 6.5O2 = 4CO2 + 5H20
Propane C3H8 + 5O2 = 3C02 + 4H2O
Butene C4H8 + 6O2 = 4CO2 + 4H20

Butane
0.75mol of C4H10

0.75 x C4 = 3mol of CO2
0.75 x H5 = 3.75mol of H2O
0.75 x 6.5O2 = 4.875mol of O2

Propane
0.10mol of C3H8

0.10 x C3 = 0.3mol of CO2
0.10 x H4 = 0.4mol of H2O
0.10 x 5O2 = 0.5mol of O2

Butene
0.15mol of C4H8

0.15 x C4 = 0.6mol of CO2
0.15 x H4 = 0.6mol of H2O
0.15 x 6O2 = 0.19mol of O2

We know, Oxygen = 6.3mol + 10% excess air = 1.1 x 6.3mol = 6.93mol.

I know I need to find the value of Nitrogen. I've seen on other threads, that they have got a value of 26.07mol...but I'm struggling to understand how they've achieved this.

This is as far as I can manage to get...Please help me with some advise!!

 

[Chestermiller]

This problem has appeared previously in PhysicsForums. Please use Search.