Ethane burns in oxygen to form CO2 and H2O according to the equation:
2C2H6 + 7O2 ⟶ 4CO2 + 6H2O.
If 1250 cc of oxygen is burnt with 300 cc of ethane.
Calculate:
(i) the volume of CO2 formed.
(ii) the volume of unused O2.
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[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 :
$$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 \text{ vol.} & : & 4 \text{ vol.} \\ 300 \text{ cc} & : & \text{x} \end{matrix}$$
∴
$x = \frac{4}{2} \times 300 = 600 \text{ cc}$
Hence, volume of carbon dioxide formed = 600 cc
(ii) To calculate the volume of unused O2 :
$$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{O}_2 \\ 2 \text{ vol.} & : & 7 \text{ vol.} \\ 300 \text{ cc} & : & \text{x} \end{matrix}$$
∴
$x = \frac{7}{2} \times 300 = 1050 \text{ cc}$
Unused oxygen = 1250 - 1050 = 200 cc.
Hence, volume of unused oxygen = 200 cc