QQuestion
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.
Exam-ready answer • 03 Mark
✓Answer
[By Lussac’s law]
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
2 vol. : 7 vol. → 4 vol.
(i) To calculate the volume of CO₂ formed:
C₂H₆ : CO₂
2 vol. : 4 vol.
80 ml : x
∴ x = (4/2) × 80 = 160 ml
Hence, volume of carbon dioxide formed = 160 ml
(ii) To calculate the volume of unused O₂:
C₂H₆ : O₂
2 vol. : 7 vol.
80 ml : x
∴ x = (7/2) × 80 = 280 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.
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