A student prepared nitrobenzene by the following reaction.
C6H6 + HNO3 → C6H5NO2 + H2O
benzene nitrobenzene
Starting with 3.9g of benzene, the student obtained 3.69g of nitrobenzene.
What is the percentage yield?
(Ar : H = 1; C = 12; N = 14; O = 16)

Answer:

22) C

Mr of benzene = (12 x 6) + (1 x 6) = 78
Mass of benzene given = 3.9 g
Moles of benzene = mass/Mr = 3.9/78 = 1/20 moles = 0.05 mol

Molar ratio of benzene and nitrobenzene = 1:1
Moles of nitrobenzene = 0.05 moles
Mr of nitrobenzene = (12 x 6) + (1 x 5) + (14 x 1) + (16 x 2) = 72 + 5 + 14 + 32 = 123 ≈ 120
Mass of nitrobenzene = moles x Mr = 0.05 x 120 = 6g

Percentage yield = (3.69 ÷ 6) x 100 = 369/6 = 61.5% ≈ 60%

Hence, C is correct.