$\therefore 0.140\, gm \frac{169}{140.5} \times 0.140$

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

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 $=210 \times 10^{-3} \,g$

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