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

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SUMMARY

This discussion focuses on calculating the composition of flue gases by volume on both a wet and dry basis, specifically addressing the combustion of butane, propane, and butene. The calculations yield 3 mol of CO2, 3.75 mol of H2O, and 4.875 mol of O2 from butane, along with similar results for propane and butene. The total oxygen requirement is determined to be 6.93 mol, leading to the calculation of nitrogen content, which is established as 26.07 mol using the air composition ratio of 21% O2 and 79% N2.

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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!!
 
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Don't know if you still need this but...

Air consists of 21% O2 and 79% N2, so 6.93mol is 21% of the total air provided.

To get the N2, divide 6.93 by 21 to get 1% then multiply by 79, which gives 26.07mol :)
 
rc404 said:
Don't know if you still need this but...

Air consists of 21% O2 and 79% N2, so 6.93mol is 21% of the total air provided.

To get the N2, divide 6.93 by 21 to get 1% then multiply by 79, which gives 26.07mol :)
Welcome to PF.

We are not allowed to do the student's work for them here at PF. Instead, ask probing questions, give hints, find mistakes, etc. The student must do the bulk of the work on their schoolwork questions.

Having said that, this thread is several months old so presumably the student no longer has this problem to solve. So in cases like this, it is okay to post a solution or an alternate solution versus the one the student came up with.
 
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