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# Combustion of Octane - equation? watch

1. Hi

Can anyone help with this question?

- Write a balanced equation for the complete combustion of octane in oxygen?

Thanks...

2. (Original post by seaspray)
Hi

Can anyone help with this question?

- Write a balanced equation for the complete combustion of octane in oxygen?

Thanks...

CnH2n+2 + (1.5n+0.5)O2 ----> nCO2 + (n+1)H2O

Therefore with n = 8

C8H18 + 12.5O2 ----> 8CO2 + 9H2O
3. (Original post by Widowmaker)
CnH2n+2 + (1.5n+0.5)O2 ----> nCO2 + (n+1)H2O

Therefore with n = 8

C8H18 + 12.5O2 ----> 8CO2 + 9H2O
Thank you so much.

How do I use that equation to calculate the volume of air (containing 20% Oxygen) needed to burn one mole of octane?

Thanks again
4. (Original post by seaspray)
Thank you so much.

How do I use that equation to calculate the volume of air (containing 20% Oxygen) needed to burn one mole of octane?

Thanks again
Mole ratio
C8H18:O2
= 1:12.5

Therefore 12.5 moles of oxygen required.

But air is 20% oxygen, so 5 x 12.5 moles of air = 60 moles

1 mole of gas occupies 24dm^3 under standard conditions.

So volume of air required = 60 x 24dm^3 = 1440dm^3
5. (Original post by Widowmaker)
Mole ratio
C8H18:O2
= 1:12.5

Therefore 12.5 moles of oxygen required.

But air is 20% oxygen, so 5 x 12.5 moles of air = 60 moles

1 mole of gas occupies 24dm^3 under standard conditions.

So volume of air required = 60 x 24dm^3 = 1440dm^3
Thanks again, but can you explain something for me?.....

Air = 20% oxygen, but where do you get the 5 from to give 60 moles?

Thanks.
6. (Original post by seaspray)
Thanks again, but can you explain something for me?.....

Air = 20% oxygen, but where do you get the 5 from to give 60 moles?

Thanks.
Air consists of 20% oxygen so you need 5 times the volume of oxygen in air to give you 12.5 moles of oxygen. I.e. 12.5 moles/60 moles x 100% = 20%. If air was 100% oxygen you'd just use 12.5 moles. I.e. 12.5/12.5 x 100% = 100% (if air contained 100% oxygen)
7. (Original post by Widowmaker)
Air consists of 20% oxygen so you need 5 times the volume of oxygen in air to give you 12.5 moles of oxygen. I.e. 12.5 moles/60 moles x 100% = 20%. If air was 100% oxygen you'd just use 12.5 moles. I.e. 12.5/12.5 x 100% = 100% (if air contained 100% oxygen)
Thank you....it's so obvious now you explained

There are two more parts to the question...

- Burning 1g of Octane produces 48.4 kJ of energy. How much heat energy is produced by burning 1g of Methane?

- Burning 1dm3 of liquid octane releases 33800kJ of energy. How much heat energy is released on burning 1dm3 of gaseous methane?

I don't even know where to start on these two

Thank you so much for your help

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Updated: February 25, 2006
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