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2017-11-09 1:09 am
The combustion of ethane produces carbon dioxide and water. If 27.6 g C2H6 are burned in the presence of excess air (O2) and 76.7 g of carbon dioxide are produced, what is the percent yield?
回答 (1)
2017-11-09 1:20 am
Molar mass of C₂H₆ = (12.0×2 + 1.0×6) g/mol = 30 g/mol
Molar mass of CO₂ = (12.0 + 16×2) g/mol = 44.0 g/mol
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Mole ratio C₂H₆ : CO₂ = 2 : 4 = 1 : 2
No. of moles of C₂H₆ burned = (27.6 g) / (30 g/mol) = 0.92 mol
Maximum no. of moles of CO₂ formed = (0.92 mol) × 2 = 1.84 mol
Theoretical yield of CO₂ = (1.84 mol) × (44.0 g/mol) = 80.96 g
Percent yield = (76.7/80.96) × 100(%) = 94.7(%)
Molar mass of CO₂ = (12.0 + 16×2) g/mol = 44.0 g/mol
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Mole ratio C₂H₆ : CO₂ = 2 : 4 = 1 : 2
No. of moles of C₂H₆ burned = (27.6 g) / (30 g/mol) = 0.92 mol
Maximum no. of moles of CO₂ formed = (0.92 mol) × 2 = 1.84 mol
Theoretical yield of CO₂ = (1.84 mol) × (44.0 g/mol) = 80.96 g
Percent yield = (76.7/80.96) × 100(%) = 94.7(%)
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