C6H6 + Br2 > C6H6Br When 30g benzene reacts with 65g of bromine, what is the theoretical yield of bromobenzene ?

Note that the molecular formula of bromobenzene is C₆H₅Br, but NOT C₆H₅Br.

Molar mass of C₆H₆ = (12.0×6 + 1.0×6) g/mol = 78.0 g/mol
Molar mass of Br₂ = 79.9 × 2 g/mol = 159.8 g/mol
Molar mass of C₆H₅Br = (12.0 × 6 + 1.0×5 + 79.9) g/mol = 156.9 g/mol

The balanced equation is :
C₆H₆ + Br₂ → C₆H₅Br + HBr

Initial number of moles of C₆H₆ = (30 g) / (78.0 g/mol) = 0.385 mol
Initial number of moles of Br₂ = (65 g) / (159.8 g/mol) = 0.407 mol

According to the equation, mole ratio C₆H₆ : Br₂ = 1 : 1
If C₆H₆ completely reacts, no. of moles of Br₂ needed = 0.385 mol < 0.407 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 reacts = 0.385 mol
Theoretically, no. of moles of C₆H₅Br produced = 0.385 mol
Theoretical yield of C₆H₅Br = (156.9 g/mol) × (0.385 mol) = 60.4 g