A mixture of methane and oxygen occupied a volume of 48 cm3. All the methane reacted and 24 cm3 of oxygen remained. What was the volume of methane present initially?

The reaction of methane ($\text{CH}_4$) with oxygen ($\text{O}_2$) produces carbon dioxide ($\text{CO}_2$) and water ($\text{H}_2\text{O}$). Since the water formed is neglected, it is considered to be in the liquid state.

$$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$$

Initial volume of oxygen = $48 \text{ cm}^3$
Unused oxygen = $24.0 \text{ cm}^3$

$$\text{Volume of O}_2 \text{ reacted} = \text{Initial volume of O}_2 - \text{Unused O}_2$$ $$\text{Volume of O}_2 \text{ reacted} = 48 \text{ cm}^3 - 24.0 \text{ cm}^3 = 24.0 \text{ cm}^3$$

From the balanced equation, the ratio of reacting gases and gaseous products by volume is:
$\text{CH}_4 : \text{O}_2 : \text{CO}_2$
$1 : 2 : 1$

Using the volume of $\text{O}_2$ reacted ($24.0 \text{ cm}^3$):

$$\text{Volume of CH}_4 \text{ reacted} = 24.0 \text{ cm}^3 \text{ O}_2 \times \frac{1 \text{ volume CH}_4}{2 \text{ volumes O}_2} = 12.0 \text{ cm}^3 \text{ CH}_4$$

This matches the initial volume of methane ($12.0 \text{ cm}^3$), confirming that all methane reacted.

$$\text{Volume of CO}_2 \text{ produced} = 24.0 \text{ cm}^3 \text{ O}_2 \times \frac{1 \text{ volume CO}_2}{2 \text{ volumes O}_2} = 12.0 \text{ cm}^3 \text{ CO}_2$$

The final volume measured is the sum of the gaseous products and any unreacted gaseous reactants.

$$\text{Final volume} = \text{Volume of CO}_2 \text{ produced} + \text{Volume of unused O}_2$$ $$\text{Final volume} = 12.0 \text{ cm}^3 + 24.0 \text{ cm}^3 = 36.0 \text{ cm}^3$$

This matches the given final volume of $36.0 \text{ cm}^3$.

The reacting gases are $\text{CH}_4$ and $\text{O}_2$. The gaseous product is $\text{CO}_2$.

$$\text{Ratio} = \text{Volume of CH}_4 \text{ reacted} : \text{Volume of O}_2 \text{ reacted} : \text{Volume of CO}_2 \text{ produced}$$ $$\text{Ratio} = 12.0 \text{ cm}^3 : 24.0 \text{ cm}^3 : 12.0 \text{ cm}^3$$

Divide by the smallest volume ($12.0 \text{ cm}^3$) to get the simplest whole number ratio:

$$\text{Ratio} = \frac{12.0}{12.0} : \frac{24.0}{12.0} : \frac{12.0}{12.0}$$ $$\text{Ratio} = 1 : 2 : 1$$

The ratio of reacting volume of methane to reacting volume of oxygen to gaseous carbon dioxide produced is $\boxed{1:2:1}$.