# Question #36044

Oct 16, 2016

The alkane is $C {H}_{4}$

#### Explanation:

General equation for combustion reaction of an alkane.

${C}_{x} {H}_{y} \left(g\right) + \left(x + \frac{y}{4}\right) {O}_{2} \to x C {O}_{2} \left(g\right) + \frac{y}{2} {H}_{2} O \left(g\right)$

By the condition of the question

$1 + x + \frac{y}{4} = x + \frac{y}{2}$

$\implies \frac{y}{2} - \frac{y}{4} = 1$

$\implies \frac{y}{4} = 1$

$\implies y = 4$

Again the hydrocarbon being an alkane $y = 2 x + 2$

$4 = 2 x + 2$

$\implies 2 x = 2$

$\implies x = 1$

So the alkane is $C {H}_{4}$ and equation becomes

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

The presence of 3 moles gas both in reactant and product side in the balanced equation satisfies our answer.