balancing out chemical equation problem?

Question

The complete combustion (reaction with oxy-
gen) of liquid octane (C8H18) a component
typical of the hydrocarbons in gasoline, pro-
duces carbon dioxide gas and water vapor.
What is the coefficient of octane in the bal-
anced equation for the reaction? (Balance
the equation with the smallest possible whole
number coefficients.)

So, I came to conclusion that the equation looks like this:

C8H18+O2 = CO2+H2O

so i solved it, and I got 1 as coefficient of C8H18. But apparently, there wasn't any choice where I could choose one, which means i probably messed it up somewhere..

could anyone tell me where I could have massed up, or the correct answer and some explanation please?

I'm really trying to learn these things...

Thanks!!! :)

Roger the Mole · Accepted Answer

C8H18+O2 = CO2+H2O
You've got 8 carbons on the left, so you need 8 carbons on the right:
C8H18 + O2 = 8 CO2 + H2O
You've got 18 hydrogens on the left, so you need 18 hydrogens on the right:
C8H18 + O2 = 8 CO2 + 9 H2O
Now you've got 25 oxygens on the right, but oxygens on the left only come in even numbers, so double everything and try again:
2 C8H18 + O2 = 16 CO2 + 18 H2O
Now you've got 50 oxygens on the right, so you need 50 oxygens on the left:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

? · Answer

i ought to furnish you some suggestion on balancing out equations generaly: >get the equation > draw containers around the seperate compounds[ this could remind you to multiply finished field by skill of the selection] >do a style of element inventory ,or checklist >attempt to observe paterns, and discover numbers to multiply with eg.fermentation of glucose; [C6H12O6] >>> [CH3CH2OH] + [CO2] ______________________ in the previous reaction > AFTER reaction C : 6 > 3 H : 12 > 6 O : 6 > 3 start to observe a patern... if we if we multiply each little thing on the suited by skill of two it rather is all equivalent! [C6H12O6] >>> 2 [CH3CH2OH] + 2[CO2]

balancing out chemical equation problem?

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