When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is obtained: C6H6+Br2→C6H5Br+HBr
A) What is the theoretical yield of bromobenzene in this reaction when 32.0 g of benzene reacts with 69.3 g of bromine?
B) If the actual yield of bromobenzene was 58.5 g , what was the percentage yield?
Chemistry question please help?
2016-10-21 12:21 am
回答 (1)
2016-10-21 3:54 pm
(A)
Molar mass of C₆H₆ = (12.0×6 + 1.0×6) g/mol = 78 g/mol
Molar mass of Br₂ = 79.9 × 2 = 159.8 g/mol
Molar mass of C₆H₅Br = (12.0×6 + 1.0×5 + 79.9) g/mol = 156.9 g/mol
C₆H₆ + Br₂ → C₆H₅Br +HBr
OR: Mole ratio C₆H₆ : Br₂ = 1 : 1
Initial no. of moles of C₆H₆ = (32.0 g) / (78.0 g/mol) = 0.410 mol
Initial no. of moles of Br₂ = (69.3 g) / (159.8 g/mol) = 0.434 mol > 0.410 mol
Hence, Br₂ is in excess, and C₆H₆ completely reacts.
According to the equation, mole ratio C₆H₆ : C₆H₅Br = 1 : 1
No. of moles of C₆H₆ reacted = 0.410 mol
Theoretically, no. of moles of C₆H₅Br produced = 0.410 mol
Theoretical yield of C₆H₅Br = (0.410 mol) × (156.9 g/mol) = 64.3 g
(B)
Percentage yield = (58.5/64.3) × 100% = 91.0%
Molar mass of C₆H₆ = (12.0×6 + 1.0×6) g/mol = 78 g/mol
Molar mass of Br₂ = 79.9 × 2 = 159.8 g/mol
Molar mass of C₆H₅Br = (12.0×6 + 1.0×5 + 79.9) g/mol = 156.9 g/mol
C₆H₆ + Br₂ → C₆H₅Br +HBr
OR: Mole ratio C₆H₆ : Br₂ = 1 : 1
Initial no. of moles of C₆H₆ = (32.0 g) / (78.0 g/mol) = 0.410 mol
Initial no. of moles of Br₂ = (69.3 g) / (159.8 g/mol) = 0.434 mol > 0.410 mol
Hence, Br₂ is in excess, and C₆H₆ completely reacts.
According to the equation, mole ratio C₆H₆ : C₆H₅Br = 1 : 1
No. of moles of C₆H₆ reacted = 0.410 mol
Theoretically, no. of moles of C₆H₅Br produced = 0.410 mol
Theoretical yield of C₆H₅Br = (0.410 mol) × (156.9 g/mol) = 64.3 g
(B)
Percentage yield = (58.5/64.3) × 100% = 91.0%
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