Question

(b) What is the percentage composition by volume of a mixture of carbon monoxide and methane, 15 ml of which required 10 ml of oxygen for complete combustion?

Answer 1

The balanced chemical equation for the given reaction is:

$$\begin{gathered} \frac{2\mathrm{CO}+{\mathrm{O}}_2\to 2{\mathrm{CO}}_2 \\ {\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}}{2\mathrm{CO}+{\mathrm{CH}}_4+3{\mathrm{O}}_2\to 3{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}} \end{gathered}$$

Therefore, two moles of carbon monoxide and one mole of methane need three moles of oxygen for complete combustion.

As we know one mole of gas occupy 22.4 L volume.
Two moles of carbon monoxide and one mole of methane occupy 44.8 litres and 22.4 litres respectively.

The ratio by volume of methane to carbon monoxide is 1:2.