How many liters of oxygen are required for the complete combustion of 5 liters of a gas mixture containing 40% methane and 60% propane?
- 15
- 20
- 21
- 25
📝 Detailed explanation
Volume CH₄ = 5 × 0.4 = 2 liters.Volume C₃H₈ = 5 × 0.6 = 3 liters.
CH₄ + 2O₂ → CO₂ + 2H₂O (Needs 2 vol O₂ per 1 vol CH₄).
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O (Needs 5 vol O₂ per 1 vol C₃H₈).
Total O₂ = (2 × 2) + (3 × 5) = 4 + 15 = 19 liters.
Wait, re-checking calculation: 4 + 15 = 19.
If option is 21, check propane volume: 3 * 5 = 15.
Calculation is 19L.