**How can you build combustion equations for both complete and incomplete combustion of hydrocarbons?**

I know there are many of you out there who are
struggling with how to build hydrocarbon combustion equations.

I think my approach is pretty fool proof so let
me know what you think.

**Can you construct complete combustion equations for hydrocarbon fuels?**

Say you are asked to construct the equation for
the complete combustion of propane: how do you do that? (Thankfully they
give you the formula of propane C

_{3}H_{8}!!!!)
Here is a routine which if you memorise it you
can’t fail to get these equations right

1
Complete combustion means just
two products water and carbon dioxide so we can write the formulas of reactants
and products like this:

C

_{3}H_{8}+ O_{2 }= CO_{2}+ H_{2}O
2
Next you can see there are
three carbon atoms in propane so that gives three CO2 molecules so the equation
is now:

C

_{3}H_{8}+ O_{2 }= 3CO_{2}+ H_{2}O
3
Next you can see there are
eight hydrogen atoms in propane but as two appear in water we'll divide eight
by two (difficult!!!) and have four water molecules on the right like this:

C

_{3}H_{8}+ O_{2 }= 3CO_{2}+ 4H_{2}O
4
All that's left is to find how
many oxygens there are on the right. I can count ten atoms so divide by
two as two atoms make an oxygen molecule and that gives us five O

_{2}on the left:
C

_{3}H_{8}+ 5O_{2 }= 3CO_{2}+ 4H_{2}O
5
The final equation then
becomes:

C

_{3}H_{8}+ 5O_{2 }= 3CO_{2}+ 4H_{2}O
This method always works for these hydrocarbon
combustion equations.

The only adjustment you'll have to make is if
the hydrocarbon has an

**even number of carbons (e.g. butane C**in it._{4}H_{10})
Then you will have to

**double every final value for each reactant and product**to be left with a whole number of oxygen molecules.
Try it and see what I mean.

**So having cracked the complete combustion equations how do you feel about the incomplete combustion equations?**

Say you are asked to construct the equation for
the

**incomplete combustion of pentane (C**into water and carbon monoxide, carbon dioxide and carbon!_{5}H_{12})
1
Let's lay the symbols down
first:

C

_{5}H_{12}+ O_{2}= C + CO + CO_{2}+ H_{2}O
2
Now things get really
interesting because there is not one straight answer.

You have five carbons in pentane and three
products that all contain carbon.

Where do you put the carbons?

We could have three carbon atoms, one carbon
monoxide and one carbon dioxide molecule.

Or we could have two carbon atoms, two carbon
monoxide molecules and one carbon dioxide molecule!!

In other words, there are several possible
equations so we have just to pick one.

So let's go with this:

C

_{5}H_{12}+ O_{2}= 3C + CO + CO_{2}+ H_{2}O
3
Next there are twelve hydrogen
atoms in pentane so dividing by two gives us six water molecules. Like this:

C

_{5}H_{12}+ O_{2}= 3C + CO + CO_{2}+ 6H_{2}O
4
Lastly, we'll count up the
oxygen atoms on the right (nine in all) and divide by two to find the number of
oxygen molecules like this:

C

_{5}H_{12}+ 4½O_{2}= 3C + CO + CO_{2}+ 6H_{2}O
5
But that's four and a half
oxygen molecules which some of us might not be too happy with as it looks
untidy (!!) so let's double up all the quantities of atoms and molecules and
we'll all feel safer like this:

2C

_{5}H_{12}+ 9O_{2}= 6C + 2CO + 2CO_{2}+ 12H_{2}O
Can you come up with some alternative equations
for this same reaction?