How do we represent the complete combustion of #"octane"#, #C_8H_18(l)#?
The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.
So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?
We need to have the same number of moles of each substance on each side. The balanced equation is:
We start out with the unbalanced equation:
We need to find the values of the coefficients
For the moment, let's leave
There are 18 moles of H on the left so we need 18 on the right, but each
Now we turn our attention to the right side. There are two O in each of 8
On the left, each
If the fraction makes you uncomfortable, another way is to multiply all the coefficients by two: