# How would you balance the combustion reaction between octane and oxygen: 2C8H18 + O2 → CO2 + H2O?

Nov 3, 2015

${C}_{8} {H}_{18} + \frac{25}{2} {O}_{2} \rightarrow 8 C {O}_{2} + 9 {H}_{2} O$

#### Explanation:

Is the above equation balanced? Don't trust my arithmetic. What would be the physical states of each participant in the reaction? Internal combustion engines rarely combust gas this cleanly, and a lot of $C$ particulate, and $C O$ gas are given off.

If (say) 1 equiv each of $C$ and $C O$ were given off by the combustion, how would I balance this reaction?

${C}_{8} {H}_{18} + x {O}_{2} \rightarrow y C {O}_{2} + C + C O + 9 {H}_{2} O$

Would the energy output be the same?