Question

3.9 g of benzene reacts with 3.3 g of $HN{O}_3$ 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:
$C_6{H}_6+HN{O}_3\to C_6{H}_{5}N{O}_2+H_{2}O$$)$

• A. 90%
• B. 50%
• C. 66%
• D. 75%

Answer 1

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 = $\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{3.9}{78}=0.05mol$

$\frac{\text{moles of benzene}}{\text{stoichiometric coefficient}}=\frac{0.05}{1}=0.05$

Mass of nitric acid = 3.3 g
Molar mass of nitric acid = 63 g/mol
Moles of nitric acid = $\frac{\text{Mass of nitric acid}}{\text{Molar mass of nitric acid}}=\frac{3.3}{63}=0.052mol$

$\frac{\text{moles of nitric acid}}{\text{stoichiometric coefficient}}=\frac{0.104}{1}=0.104$
$\therefore$Benzene is the limiting reagent.

Theoretical yield of the reaction:
$\frac{\text{moles of benzene}}{1}=\frac{\text{moles of nitrobenzene}}{1}$
$\text{Theoretical yield of nitrobenzene = 0.05 mol}$
$\text{Molar mass of nitrobenzene = 123 g/mol}$
$\text{Theoretical yield in mass = 6.15 g}$
$\text{Actual yield = 3.075 g}$
Percentage yield = $\frac{actualyield}{theoreticalyield}\times 100=\frac{3.075}{6.15}\times 100=50\%$