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Q: Combustion of Octane ( Answered,   1 Comment )
Subject: Combustion of Octane
Category: Science
Asked by: teriberry-ga
List Price: $2.00
Posted: 12 Nov 2006 12:58 PST
Expires: 12 Dec 2006 12:58 PST
Question ID: 782154
The equation for the complete combustion of octane is as follows
2C8H18 +25O2-----16CO2+H20
When 228g of octane are burnt completey. 
Calculate 1. the volume of oxygen needed
          2. the volume of carbon dioxide produced
(Assume that 1 mole of gas has a volume of 22.4dm3)
Subject: Re: Combustion of Octane
Answered By: livioflores-ga on 12 Nov 2006 13:36 PST

The equation gives the moles ratio between reactants and products. For
case, to burn 2 moles of C8H18, 25 moles of O2 are needed, 16 moles of
CO2 and 1 single mole of water are produced.

1 mole of octane weights (8*12g + 18*1g) = 114g

So 228g of octane are equal to (228/114) = 2 moles of octane; then if
the 228g of octane are burnt completely; this means that 2 moles of
octane are burnt completely; therefore 25 moles of O2 are needed.
Since each mole has a volume of 22.4dm3, the volume of oxygen needed
is (25*22.4dm3) = 560dm3

As a product of the reaction 16 moles of CO2 are produced, this amount
of CO2 has a volume of (16*22.4dm3) = 358.4dm3

I hope this helps you.


Clarification of Answer by livioflores-ga on 24 Nov 2006 17:55 PST

The commenter momomc-ga is right, the balanced equation is:
2C8H18 + 25O2 ---> 16CO2 + 18H2O

Affortunately the only part that is wrong in the original misbalanced
equation was the "+ 18H2O", specifically the number of moles of water
that the combustion produces. Since the other terms of the equation
are right and no question was made about the water, the answer does
not change.

Subject: Re: Combustion of Octane
From: momomc-ga on 24 Nov 2006 10:03 PST
The equation is not balanced.  Calculations must be based on a
balanced equation to be correct.

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