A chemistry problem by jaikirat sandhu

The answer is 3:2. 3 parts of CO.

The combustion reactions of $\ce{CO}$ and $\ce{O}$ are as follows:

$\ce{2 CO (g) + O2 (g) -> 2CO2 (g)} \\ \ce{2 H2 (g) + O2 (g) -> 2H2O} (l)$

Let the volumes (moles) of $\ce{CO}$ and $\ce{O}$ be $x$ and $y$ mL respectively. Therefore $x+y= 20$ mL. The volumes of gases before and after the combustion are as follows:

$x\ce{CO (g)} + \frac{x}{2} \ce{O2 (g) ->} x \ce{CO2 (g)} \\ y \ce{H2 (g)} + \frac{y}{2} \ce{O2 (g) -> (0)H2O} (l)$

$\begin{aligned} \Rightarrow \underbrace{x + \frac{x}{2} + y + \frac{y}{2}}_{\text{before combustion}} - \underbrace{x}_{\text{after}} & = 18 \\ \frac{3}{2}(x+y) - x & = 18 \\ \frac{3}{2} \times 20 - x & = 18 \\ 30 - x & = 18 \\ \Rightarrow x & = 30-18 = 12 \text{ mL} \\ \Rightarrow y & = 20 - 12 = 8 \text{ mL} \end{aligned}$

$\Rightarrow \dfrac{x}{y} = \dfrac{12}{8} = \dfrac{3}{2} \quad \Rightarrow a + b = 3 + 2 = \boxed{5}$