c
\(\begin{array}{lll}1 \text { mole } & 1 \text { mole } & 1 \text { mole } \\ =140.5 gm & =169 gm & =273 gm \end{array}\)

\(\therefore 0.140\, gm \frac{169}{140.5} \times 0.140\)

\(\text { L.R. } = 0.168\, gm < 0.388 \,gm\)

excess

\(\therefore\) Theoretical amount of given product formed

\(=\frac{273}{140.5} \times 0.140=0.272\, gm\)

But its actual amount formed is \(0.210\, gm\).

Hence, the percentage yield of product.

\(=\frac{0.210}{0.272} \times 100=77.20 \approx 77\)

\(OR\)

Mole of \(Ph - CoCl =\frac{0.140}{140}=10^{-3}\, mol\)

Mole of \(\begin{array}{*{20}{c}} {O\,\,\,\,\,\,\,\,\,\,} \\ {||\,\,\,\,\,\,\,\,\,\,\,} \\ {Ph - C = N{{(Ph)}_2}} \end{array}\) that should be obtained

by mol-mol analysis \(=10^{-3} \,mol\).

Theoritical mass of product \(=10^{-3} \times 273=\) \(273 \times 10^{-3} \,g\)

Observed mass of product \(=210 \times 10^{-3} \,g\)

\(=\frac{210 \times 10^{-3}}{273 \times 10^{-3}} \times 100=76.9\, \%=77\)