Chemistry - Mole Concept Question with Solution | TestHub

$1 \mathrm{mole} 1 \mathrm{mole} 1 \mathrm{mole}$

$=140.5\mathrm{gm} =169\mathrm{gm} =273\mathrm{gm}$

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

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

excess

$\therefore$ Theoretical amount of given product formed

$=\frac{273}{140.5}\times 0.140=0.272\mathrm{gm}$

But its actual amount formed is $0.210\mathrm{gm}$. Hence, the percentage yield of product.

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

OR

Mole of $\mathrm{Ph}-\mathrm{CoCl}=\frac{0.140}{140}={10}^{-3} \mathrm{mol}$

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

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

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

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