Hydrocarbons (3) How to build hydrocarbon combustion equations

This stuff is lifted from my previous post on hydrocarbon fuels because 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.

Here is what I wrote in a previous post:

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 C3H8!!!!)

1. Complete combustion means just two products water and carbon dioxide so we can write the formulas of reactants and products like this:                                                                             C3H8        +          O2              =          CO2      +   H2O    
2. Next you can see there are three carbon atoms in propane so that gives three CO2 molecules so the equation is now:        C3H8        +          O2              =          3CO2      +   H2O   
3. Next you can see there are eight hydrogen atoms in propane so as two appear in water we'll divide eight by two (difficult!!!) and have four water molecules on the right like this:        C3H8        +          O2              =          3CO2      +   4H2O  
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 O2 on the left: C3H8        +          5O2              =          3CO2      +   4H2O 
5. Our finished equation is:                                                                                                                  C3H8        +          5O2              =          3CO2      +   4H2O 
6. 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 C4H10) in it when 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 (C5H12) into water and carbon monoxide, carbon dioxide and carbon!

1. Let's lay the symbols down first:                                                                                                       C5H12         +           O2       =         C            +          CO         +          CO2         +    H2O  
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:                                                                                                                             C5H12         +           O2       =         3C            +          CO         +          CO2         +    H2O
3. Next there are twelve hydrogen atoms in pentane so dividing by two gives us six water molecules.  Like this:                                                                                                                 C5H12         +           O2       =         3C            +          CO         +          CO2         +    6H2O
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:                                                                                          C5H12         +          41/2 O2       =         3C            +          CO         +          CO2      +    6H2O
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:                                                                                                                                       2C5H12         +        9O2       =         6C            +          2CO         +          2CO2         +    12H2O

Can you come up with some alternative equations for this same reaction?

Pages on the "Mole" and "Using the Mole" in chemical calculations are here