Combustion of 50 cm3 of carbon(II)oxide in 100 cm3 of oxygen. Calculate the volume of carbon(IV)oxide produced.

The combustion of carbon(II)oxide (CO) in oxygen (O$_2$) is given by the equation:

Initial volume of CO = $50 \text{ cm}^3$
Initial volume of O$_2$ = $100 \text{ cm}^3$

According to Gay-Lussac's Law of Gaseous Volumes, the volumes of gaseous reactants and products are in the same ratio as their stoichiometric coefficients in the balanced equation (at constant temperature and pressure).
From the equation, 2 volumes of CO react with 1 volume of O$_2$ to produce 2 volumes of CO$_2$.
The ratio CO : O$_2$ : CO$_2$ is $2 : 1 : 2$.

Let's determine how much O$_2$ is needed to react completely with $50 \text{ cm}^3$ of CO:
Volume of O$_2$ needed = $50 \text{ cm}^3 \text{ CO} \times \frac{1 \text{ volume O}_2}{2 \text{ volumes CO}} = 25 \text{ cm}^3 \text{ O}_2$

Since we have $100 \text{ cm}^3$ of O$_2$ available, and only $25 \text{ cm}^3$ is needed, O$_2$ is in excess.
Therefore, CO is the limiting reactant.

Since CO is the limiting reactant, the amount of product (CO$_2$) formed is determined by the initial amount of CO.
From the stoichiometric ratio, 2 volumes of CO produce 2 volumes of CO$_2$.
Volume of CO$_2$ produced = $50 \text{ cm}^3 \text{ CO} \times \frac{2 \text{ volumes CO}_2}{2 \text{ volumes CO}} = 50 \text{ cm}^3 \text{ CO}_2$

Initial volume of O$_2$ = $100 \text{ cm}^3$
Volume of O$_2$ reacted = $25 \text{ cm}^3$
Volume of O$_2$ remaining = $100 \text{ cm}^3 - 25 \text{ cm}^3 = 75 \text{ cm}^3$

The resulting mixture will consist of the product (CO$_2$) and the unreacted excess reactant (O$_2$).
Total volume of resulting mixture = Volume of CO$_2$ produced + Volume of O$_2$ remaining
Total volume = $50 \text{ cm}^3 + 75 \text{ cm}^3 = 125 \text{ cm}^3$

The final answer is $\text{125 cm}^3$.