Consider the below-given reaction. The percentage yield of an amide product is ______. (Round off to the Nearest Integer).
[Given: Atomic mass: $C$ $= 12.0 u$ , $H$ $= 1.0 u$ , $N$ $= 14.0 u$ , $O$ $= 16.0 u$ , $Cl$ $= 35.5 u$ ]

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Solution

Step 1: The mass of reactants and product used in the reaction is given in the equation as follows:
Step 2: First let us determine the limiting reagent in the reaction.
We know that $140.5 g$ of $Ph - CO - Cl$ reacting with $169 g$ of $Ph - NH - Ph$ [ $140 g$ and $169 g$ are the respective molecular weights of the compounds]
By unitary method $1 g$ $Ph - CO - Cl$ $= \frac{169}{140.5} g$ $Ph - NH - Ph$
So, $140 g$ of $Ph - CO - Cl$ $= \frac{169}{140.5} \times 140 g$ $Ph - NH - Ph$
$= 0.168 g$
Now, the observed mass of $Ph - NH - Ph$ i.e $0.168 g$ is less than the theoretical mass i.e $0.388 g$ .
So, the amount of $Ph - CO - Cl$ is limited in reaction. Hence the reaction will proceed according to the mass of $Ph - CO - Cl$ .
Now, $140.5 g$ of $Ph - CO - Cl$ reacting to form $= 273 g$ of $Ph - CO - NH - {(Ph)}_{2}$
So, $1 g$ $Ph - CO - Cl$ reacting to form $= \frac{273}{140.5} g$ of $Ph - CO - NH - {(Ph)}_{2}$
$0.140 g$ of $Ph - CO - Cl$ reacting to form $= \frac{273}{140.5} \times 0.140 g$ of $Ph - CO - NH - {(Ph)}_{2}$
$= 0.272 g$
The observed mass of product $Ph - CO - NH - {(Ph)}_{2}$ obtained $= 0.272 g$
The theoretical mass of product $Ph - CO - NH - {(Ph)}_{2}$ obtained $= 0.210 g$
Step 3: The percentage yield of $Ph - CO - NH - {(Ph)}_{2}$ $= \frac{T h e o r e t i c a l m a s s}{o b s e r v e d m a s s} \times 100$
$= \frac{0.210}{0.272} \times 100 = 77 %$

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Consider the below-given reaction. The percentage yield of an amide product is ______. (Round off to the Nearest Integer). [Given: Atomic mass: C=12.0u, H=1.0u, N=14.0u, O=16.0u, Cl=35.5u]

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