Solve the following numerical problem.

Ethane burns in oxygen according to the chemical equation:

2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

If 80 ml of ethane is burnt in 300 ml of oxygen, find the composition of the resultant gaseous mixture when measured at room temperature.

[By Lussac's law]

$\begin{matrix} 2\text{C}$2\text{H}6 & + & 7\text{O}2 &\longrightarrow & 4\text{CO}2 & + & 6\text{H}_2\text{O} \ 2\text{ vol.} & : & 7 \text{ vol.} & \longrightarrow & 4\text{ vol.} \end{matrix}

(i) To calculate the volume of CO₂ formed :

$\begin{matrix}\text{C}$2\text{H}6 & : & \text{CO}_2 \ 2 \text{ vol.} & : & 4 \text{ vol.} \ 80 \text{ ml} & : & \text{x} \end{matrix}

$\therefore x = \dfrac{4}{2} \times 80 = 160 \text{ ml}$

Hence, volume of carbon dioxide formed = 160 ml

(ii) To calculate the volume of unused O₂ :

$\begin{matrix}\text{C}$2\text{H}6 & : & \text{O}_2 \ 2 \text{ vol.} & : & 7 \text{ vol.} \ 80 \text{ ml} & : & \text{x} \end{matrix}

$\therefore x = \dfrac{7}{2} \times 80 = 280 \text{ ml}$

Unused oxygen = 300 - 280 = 20 ml.

Hence, volume of unused oxygen = 20 ml

Therefore, the resultant gaseous mixture consists of 160 ml of carbon dioxide and 20 ml of unused oxygen.