[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 :
$$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 \text{ vol.} & : & 4 \text{ vol.} \\ 80 \text{ ml} & : & \text{x} \end{matrix}$$
$\therefore x = \frac{4}{2} \times 80 = 160 \text{ ml}$
Hence, volume of carbon dioxide formed = 160 ml
(ii) To calculate the volume of unused O2 :
$$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{O}_2 \\ 2 \text{ vol.} & : & 7 \text{ vol.} \\ 80 \text{ ml} & : & \text{x} \end{matrix}$$
$\therefore x = \frac{7}{2} \times 80 = 280 \text{ ml}$
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.