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Question

3.9 g of benzene reacts with 3.3 g of HNO3 to produce 3.075 g of nitrobenzene.
(atomic mass of H = 1 u, N = 14 u C = 12 u and O = 16 u)
The percent yield of the reaction is:

(The balanced chemical equation is:
C6H6+HNO3C6H5NO2+H2O)

A
90%
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B
50%
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C
66%
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D
75%
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Solution

The correct option is B 50%
From the reaction, 1 mole of benzene gives 1 mole of nitrobenzene.

Finding the limiting reagent:
Mass of benzene = 3.9 g
Molar mass of benzene = 78 g/mol
Moles of benzene = Mass of benzeneMolar mass of benzene=3.978=0.05 mol

moles of benzenestoichiometric coefficient=0.051=0.05

Mass of nitric acid = 3.3 g
Molar mass of nitric acid = 63 g/mol
Moles of nitric acid = Mass of nitric acidMolar mass of nitric acid=3.363=0.052 mol

moles of nitric acidstoichiometric coefficient=0.1041=0.104
Benzene is the limiting reagent.

Theoretical yield of the reaction:
moles of benzene1=moles of nitrobenzene1
Theoretical yield of nitrobenzene = 0.05 mol
Molar mass of nitrobenzene = 123 g/mol
Theoretical yield in mass = 6.15 g
Actual yield = 3.075 g
Percentage yield = actual yieldtheoretical yield×100=3.0756.15×100=50%

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