1. Balanced Chemical Equation:
The reaction of benzene with bromine to produce bromobenzene is:
C₆H₆ + Br₂ → C₆H₅Br + HBr
2. Molar Masses:
* Benzene (C₆H₆): 78.11 g/mol
* Bromine (Br₂): 159.80 g/mol
* Bromobenzene (C₆H₅Br): 157.01 g/mol
3. Determine the Limiting Reactant:
* Calculate moles of benzene: (26.0 g) / (78.11 g/mol) = 0.333 mol
* Calculate moles of bromine: (56.3 g) / (159.80 g/mol) = 0.352 mol
The balanced equation shows a 1:1 mole ratio between benzene and bromine. Since we have slightly more moles of bromine, benzene is the limiting reactant.
4. Calculate Theoretical Yield:
* Use the mole ratio from the balanced equation: 1 mol benzene produces 1 mol bromobenzene.
* Calculate moles of bromobenzene produced: 0.333 mol benzene * (1 mol bromobenzene / 1 mol benzene) = 0.333 mol bromobenzene
* Convert moles of bromobenzene to grams: (0.333 mol) * (157.01 g/mol) = 52.3 g bromobenzene
Therefore, the theoretical yield of bromobenzene is 52.3 grams.
