How do we represent the complete combustion of octane?

Apr 10, 2016

${C}_{8} {H}_{18} \left(g\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(g\right)$

Balance in the order: $C$; $H$; $O$.

Explanation:

Balance the carbons, then the hydrogens, then finally the oxygens. If you like you can double the entire reaction, to remove the non-stoichiometric coefficient on dioxygen. There is no need to do so.

Is this equation balanced? Does garbage in equal garbage out? If you made a cash transaction with a large bank note, would you immediately know whether you had been short-changed or not? I think you would know, and this is the basis of stoichiometry. You have to try to develop this facility with chemical reactions, which are (MUCH much larger) atomic and molecular transactions rather than cash transactions.

Try it with nonane, ${C}_{9} {H}_{20}$.