# Question #a32cf

Mar 11, 2016

See below

#### Explanation:

$2 {C}_{8} {H}_{18} + 25 {O}_{2} = 16 C {O}_{2} + 18 {H}_{2} O$

you can remember following also
${C}_{x} {H}_{y} + \left(x + \frac{y}{4}\right) {O}_{2} = x C {O}_{2} + \frac{y}{2} {H}_{2} O$

$\implies 2 {C}_{x} {H}_{y} + \left(2 x + \frac{y}{2}\right) {O}_{2} = 2 x C {O}_{2} + y {H}_{2} O$

Oxidation Number Method
${C}_{8} {H}_{18} + {O}_{2} \to C {O}_{2} + {H}_{2} O$
change in oxidation no of C $\implies - \frac{9}{4} \to + 4 = \frac{25}{4} u n i t$ oxidation
change in oxidation no of O $\implies 0 \to - 2 = 2 u n i t$ Reduction
ratio of no of C and O =$2 : \frac{25}{4} = 8 : 25$
ratio of no of ${C}_{8} H 18$ and O$= 1 : 25$
ratio of no of ${C}_{8} H 18$ and O$= 2 : 50$
ratio of no of ${C}_{8} H 18$ and${O}_{2} = 2 : 25$