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  • Consider the above reaction. The percentage yield of amide product is __________. (Round off to the Nearest Integer). (Given : Atom

Consider the above reaction. The percentage yield of amide product is __________. (Round off to the Nearest Integer). (Given : Atomic mass : C : 12.0 u, H : 1.0u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
 

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Finding Limiting reagent-

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

\text { Mole of } \mathrm{Ph}_2\mathrm{NH}=\frac{0.388}{169}=2 \times 10^{-3} \mathrm{~mol}

So, reactant Ph-CoCl is a limiting reagent.

Hence, the Mole of Ph-CoCl reactant and product PhCON(Ph)2 will be the same, in theoritical.

So, the theoretical mole of PhCON(Ph)2  will be 10-3 mol.

Theoretical mass of product = 10-3 × 273 = 273 × 10-3 g

Observed mass of product= 210 × 10-3 g

The percentage yield of amide product is:

\% \text { Yield }=\frac{\text { Actual Yield }}{\text { Theoretical Yield }} \times 100 \%

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

\text { \%yield of product }=76.9 \%=77 \%

Ans = 77

Posted by

Kuldeep Maurya

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