Question

$66{\text{cm}}^3$ of oxygen was mixed with $44{\text{cm}}^3$ of carbon monoxide and the mixture ignited. Calculate the volume of oxygen used up and the volume of carbon dioxide so formed.

Answer 1

Given information:

• $66{\text{cm}}^3$ of oxygen $\left({\text{O}}_2\right)$
• $44{\text{cm}}^3$ of carbon monoxide$\left(\mathrm{CO}\right)$

Balanced chemical reaction

• Carbon monoxide reacts with oxygen and results in the formation of carbon dioxide $\left({\mathrm{CO}}_2\right)$.

$2\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Gay-Lussac's Law of combining volumes

• According to this law, when gases react to form new gases, the ratio of the volume of gaseous reactants to the gaseous products is expressed as a simple whole-number ratio. The conditions required are constant pressure and temperature. This law is only applicable to gases. As a result, the volumes of solids and liquids are assumed to be zero.
• The balanced chemical reaction in terms of volume of reactants and products is as follows:

$$\begin{gathered} 2\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right) \\ 2\mathrm{vols}1\mathrm{vol}2\mathrm{vols} \end{gathered}$$

Calculation of the volume of Oxygen used up

• From the balanced equation, it is clear that $2$ volumes of Carbon monoxide require $1$ volume of oxygen.
• Therefore, $44{\text{cm}}^3$ Carbon monoxide will require $\frac{1}{2}\times 44{\text{cm}}^3{\text{=22 cm}}^3$ volume of Oxygen.

Calculation of the volume of Carbon dioxide formed:

• From the balanced equation, it is clear that $2$ volumes of Carbon monoxide produce $2$ volumes of Carbon dioxide.
• Therefore, $44{\text{cm}}^3$ Carbon monoxide will produce $\frac{2}{2}\times 44{\text{cm}}^3{\text{=44 cm}}^{\text{3}}$ volume of Carbon dioxide.

Hence, $22{\text{cm}}^3$ volumes of Oxygen are used up, and $44{\text{cm}}^3$ volumes of Carbon dioxide are formed.