# How do you balance _C4H16+_O2--> _CO2+_H2O? I have tried to do balance them but i can&#39;t..this is very tricky... thanks..?

Oct 21, 2015

Combustion of butane:
${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(g\right)$

#### Explanation:

Hydrocarbons (of all stamps) react with oxygen to give $C {O}_{2}$ and ${H}_{2} O$. Sometimes the carbon in hydrocarbons, where oxygen is limited, react to give particulate $C$, and gaseous $C O$ (this occurs particularly in diesel engines). Balance the carbons (as $C {O}_{2}$ for complete combustion), then the hydrogens (as water), and then the oxygen reactant (it doesn't matter if I have a half coefficient, because I could remove it by doubling the entire reaction:

$2 {C}_{4} {H}_{10} \left(g\right) + 13 {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 10 {H}_{2} O \left(g\right)$

For every reactant particle is there a corresponding product particle? Hmm, $8 C ' s$, $20 H ' s$, $26 O ' s$. So this equation is indeed stoichiometrically balanced, as required.

You should try this routine with other hydrocarbons, $C {H}_{4}$, ${H}_{3} C - C {H}_{3}$, hexane, kerosene, diesel. These questions invariably appear on A level exams, and with a little bit of preparation they are trivial exercises. Good luck!