Ethane burns in O₂ to form CO₂ and H₂O

2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

If 1250 cc. of oxygen is burnt with 300 cc of ethane.

Calculate :

(i) volume of CO₂ formed.

(ii) volume of unused O₂.

[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.} \ 300 \text{ cc} & : & \text{x} \end{matrix}

∴

$x = \dfrac{4}{2} \times 300 = 600 \text{ cc}$

Hence, volume of carbon dioxide formed = 600 cc

(ii) (i) To calculate the volume of unused O₂ :

$\begin{matrix}\text{C}$2\text{H}6 & : & \text{O}_2 \ 2 \text{ vol.} & : & 7 \text{ vol.} \ 300 \text{ cc} & : & \text{x} \end{matrix}

∴

$x = \dfrac{7}{2} \times 300 = 1050 \text{ cc}$

Unused oxygen = 1250 - 1050 = 200 cc.

Hence, volume of unused oxygen = 200 cc