Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 7

[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