Consider the above reaction. The percentage yield of amide product...

$1mole1mole1mole$

$= 140 . 5gm = 169gm = 273gm$

$∴ 0 . 140gm\dfrac{169}{140 . 5} \times 0 . 140$

$L . R . = 0 . 168gm < 0 . 388gm$

excess

$∴$ Theoretical amount of given product formed

$= \dfrac{273}{140 . 5} \times 0 . 140 = 0 . 272gm$

But its actual amount formed is $0 . 210gm$. Hence, the percentage yield of product.

$= \dfrac{0 . 210}{0 . 272} \times 100 = 77 . 20 \approx 77$

OR

Mole of $Ph - CoCl = \dfrac{0 . 140}{140} = {10}^{- 3}mol$

by mol-mol analysis $= {10}^{- 3}$ mol.

Theoritical mass of product $= {10}^{- 3} \times 273 =$ $273 \times {10}^{- 3}g$

Observed mass of product $= 210 \times {10}^{- 3}g$

$$ yield of product $= \dfrac{210 \times {10}^{- 3}}{273 \times {10}^{- 3}} \times 100 = 76 . 9$