ICSE Class 10 Chemistry Question 18 of 23

Model Paper 2 — Question 14

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14
Question

Question 6(ii)

1250 cc of oxygen was burnt with 300 cc of ethane (C2H6). Calculate the volume of the unused oxygen and the volume of the carbon dioxide formed:

2C2H6 + 7O2 ⟶ 4CO2 + 6H2O

Answer

2C2H6+7O24CO2+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}

[By Gay Lussac's law]

2 Vol. of C2H6 requires 7 Vol. of oxygen

∴ 300 cc C2H6 will require 72\dfrac{7}{2} x 300

= 1050 cc of Oxygen

Hence, unused oxygen = 1250 - 1050 = 200 cc

Similarly,

2 Vol. of C2H6 produces 4 Vol. of carbon dioxide

∴ 300 cc C2H6 produces 42\dfrac{4}{2} x 300

= 600 cc of Carbon dioxide

Hence, carbon dioxide produced = 600 cc.