Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. If the ratio of carbon dioxide to carbon monoxide formed is 9:1 when x cm^3 of ethene is burnt, what is the volume of oxygen gas consumed in the reaction?

1 Answer
May 29, 2017

Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. The balanced equation of this incomplete combustion reaction is as follows.

${C}_{2} {H}_{4} \left(g\right) + \frac{5}{2} {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + C O \left(g\right) + 2 {H}_{2} O \left(g\right)$

But as per question the ratio of volumes of $C {O}_{2} \left(g\right) : C O \left(g\right)$ produced is $9 : 1$. So this also represents the mole ratio of
$C {O}_{2} \left(g\right) : C O \left(g\right)$ produced.

Hence adjusting this ratio balanced equation may be written as follows
$10 {C}_{2} {H}_{4} \left(g\right) + \frac{39}{2} {O}_{2} \left(g\right) \to 9 C {O}_{2} \left(g\right) + C O \left(g\right) + 20 {H}_{2} O \left(g\right)$

Here we see $10 c {m}^{3}$ ${C}_{2} {H}_{4} \left(g\right)$ will require $\frac{39}{2} = 19.5 c {m}^{3} \text{ } {O}_{2} \left(g\right)$

So to burn $x c {m}^{3}$ ethene the required volume of oxygen will be $\frac{19.5}{10} \times x c {m}^{3} = 1.95 x c {m}^{3}$