A mixture of 10 liters of methane and ethane has a density of 1.071 g/l at STP. Calculate the volume of oxygen required for the complete combustion of this mixture.
- 25 liters
- 20 liters
- 30 liters
- 35 liters
📝 Detailed explanation
M_avg = 1.071 × 22.4 = 24 g/mol.Let x = moles CH₄, y = moles C₂H₆.
(16x + 30y) / (x + y) = 24.
16x + 30y = 24x + 24y.
6y = 8x → y/x = 4/3.
In 10 L mixture: V(CH₄) = 10 × (3/7) ≈ 4.28 L; V(C₂H₆) = 10 × (4/7) ≈ 5.72 L.
O₂ for CH₄ = 2 × V(CH₄) = 8.56 L.
O₂ for C₂H₆ = 3.5 × V(C₂H₆) = 20.02 L.
Total O₂ = 8.56 + 20.02 = 28.58 L.
Re-checking calculation with M=24: (3*16 + 4*30)/7 = (48 + 120)/7 = 168/7 = 24.
O₂ = (3*2 + 4*3.5)/7 * 10 = (6 + 14)/7 * 10 = 200/7 ≈ 28.5 L.
Closest option is 25 or 30 depending on rounding.