Consider the above reaction. The percentage yield of amide product is . (Given: Atomic mass: C: 12.0 u, H: 1.0 u, N: 14.0 u, 0: 16.0 u, cl: 35.5 u )

$140.5 g$ of $\mathrm{Ph}-\mathrm{CO}-\mathrm{Cl}$ reacts with $169 g$ of  $\mathrm{Ph}-\mathrm{NH}-\mathrm{Ph}$

So, $0.140 g$ of $\mathrm{Ph}-\mathrm{CO}-\mathrm{Cl}$ reacts with$=\frac{169}{140.5}\times 0.140 g$

$=0.168 g$

But we have $0.388 \mathrm{g}$ of $\mathrm{P} \mathrm{h}-\mathrm{N} \mathrm{H}-\mathrm{P} \mathrm{h}$.

Thus, $\mathrm{Ph}-\mathrm{CO}-\mathrm{Cl}$ is the limiting reagent.

Now,

140.5g of $\mathrm{Ph}-\mathrm{CO}-\mathrm{Cl}$ reacts to give $273 \mathrm{g}$ of $\mathrm{Ph}-\mathrm{CO}-\mathrm{N}-(\mathrm{Ph}{)}_2$

$\Rightarrow 0.140 \mathrm{g}$ of gives $=\frac{273}{140.5}\times 0.140 g$

$=0.272 \mathrm{g}$

Since, $\%$ yield $=\frac{\mathrm{Experimentalvalue}}{\mathrm{Theoreticalvalue}}\times 100$

$$\begin{gathered} =\frac{0.210}{0.272}\times 100 \\ =77\% \end{gathered}$$

Hence, the correct answer is $77 \%$