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ICSE Class 10 Chemistry Question 14 of 24

Solved 2023 Question Paper ICSE Class 10 Chemistry — Question 7

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7
Question

Question 4(iii)

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.

Answer

[By Lussac's law]

2C2H6+7O2⟶4CO2+6H2O2 vol.:7 vol.⟶4 vol.\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 :

C2H6:CO22 vol.:4 vol.80 ml:x\begin{matrix}\text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 \text{ vol.} & : & 4 \text{ vol.} \\ 80 \text{ ml} & : & \text{x} \end{matrix}

∴x=42×80=160 ml\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 :

C2H6:O22 vol.:7 vol.80 ml:x\begin{matrix}\text{C}_2\text{H}_6 & : & \text{O}_2 \\ 2 \text{ vol.} & : & 7 \text{ vol.} \\ 80 \text{ ml} & : & \text{x} \end{matrix}

∴x=72×80=280 ml\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.