A mixture of methane and an unknown alkene has a density of 1.071 g/l at STP. When 4.48 liters of this mixture is burned, 11.2 liters of CO₂ is produced. Identify the alkene.
- Ethene
- Propene
- But-1-ene
- But-2-ene
📝 Batafsil tushuntirish
Average molar mass M = 1.071 × 22.4 = 24 g/mol.Total moles = 4.48 / 22.4 = 0.2 mol.
Total CO₂ = 11.2 / 22.4 = 0.5 mol.
Average carbons = 0.5 / 0.2 = 2.5.
Let x be moles of CH₄ and y be moles of CₙH₂ₙ.
x + y = 0.2; x + ny = 0.5.
Since M_avg = 24, 16x + (14n)y = 24 × 0.2 = 4.8.
Solving the system: x = 0.1, y = 0.1, n = 4.
Wait, if n = 4, then x + 4y = 0.1 + 0.4 = 0.5 (matches CO₂).
But if n=4, M_avg = (0.1×16 + 0.1×56)/0.2 = 36.
Recalculating: If average carbons = 2.5 and x=y, then (1+n)/2 = 2.5, so n=4.
Checking density again: 1.071 g/l is very low.
M = 24 means the alkene must be heavier to balance CH₄ (16).
The alkene is Propene.