Calculate the actual fuel:air ratio by volume and by mass

[PCal]

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. Data: Net calorific value (MJ m–3) at 25ºC of: Butane (C4H10) = 111.7 MJ m–3 Butene (C4H8) = 105.2 MJ m–3 Propane (C3H8) = 85.8 MJ m–3 Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and 76.7% nitrogen by mass. Atomic mass of C = 12, O = 16, N=14 and H = 1.
Calculate actual fuel:air ratio by volume and by mass
I've done this a few times and got different answers every time, I think I might be correct with this one but I'm not sure so any help would be appreciated!

Relevant Equations

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

By Volume:
Butane 0.75x6.5= 4.875 moles of O2
Propane 0.1x5= 0.5 moles of O2
Butene 0.15x6= 0.9 moles of O2
Total O2 4.875+0.5+9= 6.275 moles
Air required 6.275/0.21=29.88 moles/m^3
With 10% excess air 29.88x1.1= 32.868moles
1 mole fuel:32.87 moles of air

By Mass:
CO2 Moles (0.75x4)+(0.1x3)+(0.15x4)= 3.9
H2 Moles (0.75x10)+(0.1x8)+(0.15x8)= 9.5
O2 Moles (0.75x13)+(0.1x10)+(0.15x12)= 12.55
Nitrogen (12.55/21)x79= 47.21

Moles x Molecular mass
CO2 3.9x12= 46.8
H2 9.5x2= 19
O2 12.55x32= 401.6
Nitrogen= (401.6/23.3)x76.7=1322

Total Fuel= 46.8+19=65.8
Total Air= 401.6+1322=1723.6 +10% excess= 1895.96

1895.96/65.8= 1 mole fuel:28.81 mole of air

 

[Chestermiller]

https://www.physicsforums.com/threads/net-calorific-value.888755/#post-5590759

 

[PCal]

Sorry I'm confused, I'm struggling to see anything in that thread that helps with calculating the ratios? It looks like that's for the next part of my question.

 

[Chestermiller]

Your by-volume analysis looks OK. You took as a basis one mole of the fuel gas, and then determined the molar amounts of oxygen and nitrogen required. This was the correct thing to do.

But the by-mass analysis does not look OK. All you need to do is convert the moles from the by-volume analysis to mass for the by-mass analysis. This does not involve the CO2. Just multiply the numbers of moles of each species in part 1 by their molecular weights.

 

[PCal]

Oh amazing so...
Butane 4.875 x 58= 282.75g
Propane 0.5 moles x 44 = 22g
Butene 0.9 x 56 = 50.4g
Total mass= 355.15g

1895.96/355.15 = 5.33
= 1 : 5.33
Does this look better please?

 

[Chestermiller]

You have 0.75 moles of butane having a mass of 43.5 grams.

 

[PCal]

Aaa I think I understand. So...
Butane 0.75 x 58 = 43.52g
Propane 0.1 x 44 = 4.4g
Butene 0.15 x 8.4g
Total mass= 56.32g
1895.96/56.32 = 33.664
= 1 : 33.67

 

[Chestermiller]

Aaa I think I understand. So...
Butane 0.75 x 58 = 43.52g
Propane 0.1 x 44 = 4.4g
Butene 0.15 x 8.4g
Total mass= 56.32g
1895.96/56.32 = 33.664
= 1 : 33.67

32.97 moles of air has a mass of about 950 grams.

 

[PCal]

Butane 0.75 x 58 = 43.5g
Propane 0.1 x 44 = 4.4g
Butene 0.15 x 8.4g
Total mass= 56.3g

0.75 x 208 = 156g
0.1 x 160 = 16g
0.15 x 192 = 28.8g
Total O2 =200.8

(200.8/23.3) x 76.7 = 661

(200.8 + 661) x 1.1 = 947.97
947.97/ 56.3 = 16.83
1: 16.83
Please say this is correct?

 

[Chestermiller]

OK. This is correct.

 

[PCal]

OK. This is correct.

You haven’t said that because I asked you to?? Haha I’m currently doing a happy dance