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?

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 $\frac{19.5}{10} \times x c {m}^{3} = 1.95 x c {m}^{3}$