# What are the the product(s) of the complete combustion of any hydrocarbon?

Nov 9, 2016

$\text{Hydrocarbon + oxygen "rarr" Carbon dioxide and water (+ energy)}$

#### Explanation:

You can write out general formula for combustion of any alkane, ${C}_{n} {H}_{2 n + 2}$:

${C}_{n} {H}_{2 n + 2} + \frac{3 n + 1}{2} {O}_{2} \rightarrow n C {O}_{2} + \left(n + 1\right) {H}_{2} O$

And for any olefin:

${C}_{n} {H}_{2 n} + \frac{3 n}{2} {O}_{2} \left(g\right) \rightarrow n C {O}_{2} \left(g\right) + n {H}_{2} O$

You will have to see if it works for the simpler alkanes and olefins. I would not recommend ever memorizing this formula. However, I would practise representing the combustion reaction with hexanes, or propylene, or other hydrocarbons. You should be able to do it quickly and reliably. All you are doing is balancing mass and charge. Energy is also balanced as well as mass and charge. What do I mean by this?

Sometimes (in fact most of the time), incomplete combustion occurs to give carbon monoxide and carbon (as soot) as the products of incomplete combustion. Benzene usually combusts very incompletely, for which the sooty flame is evidence:

${C}_{6} {H}_{6} \left(l\right) + 6 {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + C \left(s\right) + C O \left(g\right) + 3 {H}_{2} O \left(l\right)$

Of course I don't know that 12 g of carbon or 28 g of carbon monoxide resulted from the combustion of 72 g of benzene, but I can certainly conceive of, and represent the possibility by this stoichiometry. Can you try to represent the complete combustion of benzene: the products are carbon dioxide and water as before?