Question

How do we represent the complete combustion of `"octane"`, `C_8H_18(l)`?

Answer 1

`C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)`

Explanation:

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?

Answer 2

We need to have the same number of moles of each substance on each side. The balanced equation is:

`C_8H_18 + 25/2` `O_2 = 8` `CO_2 + 9` `H_2O` or `2` `C_8H_18 + 25` `O_2 = 16` `CO_2 + 18` `H_2O`

Explanation:

We start out with the unbalanced equation:

`a` `C_8H_18 + b` `O_2 = c` `CO_2 + d` `H_2O`

We need to find the values of the coefficients `a, b, c` and `d` to balance the equation.

For the moment, let's leave `a` as 1. There are 8 moles of C on the left, so to balance we need 8 moles of C on the right, so let's make `c = 8`, because each `CO_2` contains 1.

`C_8H_18 + b` `O_2 = 8` `CO_2 + d` `H_2O`

There are 18 moles of H on the left so we need 18 on the right, but each `H_2O` contains 2, so we make `d = 9`.

`C_8H_18 + b` `O_2 = 8` `CO_2 + 9` `H_2O`

Now we turn our attention to the right side. There are two O in each of 8 `CO_2` for a total of 16 in the carbon dioxide and one in each of 9 `H_2O` for a total of 9 in the water, so we need 25 O all together.

On the left, each `O_2` contains two O, so one way to balance the equation is to take `25/2` of them:

`C_8H_18 +` `25/2 O_2 =` `8` `CO_2 + 9` `H_2O`

If the fraction makes you uncomfortable, another way is to multiply all the coefficients by two:

`2` `C_8H_18 + 25` `O_2 = 16` `CO_2 + 18` `H_2O`