Question

Write a balanced chemical equation for the complete combustion (in the presence of excess oxygen) of: How do you start? (a) C12H26 (a component of kerosene) (b)

Answer 1

`C_12H_26+18(1)/(2)O_2rarr12CO_2+13H_2O`

Start with the C atoms. There are 12 on the left so that must form 12`CO_2` on the right.

Then balance the `H` atoms. There are 26 on the left so that must form `13H_2O` on the right.

Always leave the `O` atoms last.

Check how many on the right.

There are 12 x 2 = 24 plus 13 x 1 = 37 in total.

So we need 37 `O` atoms on the left.

Oxygen arrives as `O_2` molecules so to provide these 37 `O` atoms we need 37/2 = `18(1)/(2)` `O_2` molecules.

This gives us the equation.

nb. `18(1)/(2)` does not refer to 18.5 molecules, which you cannot have.

It refers to 18.5 moles of oxygen molecules.

You could easily x all the species by 2 to give:

`2C_12H_26+37O_(2)rarr24CO_2+26H_2O`

An equation is just a statement of the relative number of moles which are reacting.