Question

Benzamide can be prepared by the action of concentrated ammonia upon benzoyl chloride.
$\underset{Benzoylchloride}{C_6{H}_{3}COCl}+2N{H}_3\to \underset{Benzamide}{C_6{H}_{5}CON{H}_2}+N{H}_{4}Cl$
In one such experiment, 65 cc of concentrated ammonia (in excess) was reacted with 15 g of benzoyl chloride to give 11.1 g of pure benzamide. Molar masses: benzoyl chloride (141); benzamide (121). The percentage yield of benzamide is:

• A. $\frac{11.1}{15}\times 100$
• B. $\frac{(15-11.1)}{15}\times 100$
• C. $\frac{11.1}{65}\times 100$
• D. $\frac{121}{141}\times 100$
• E. $\frac{11.1\times 141}{121\times 15}\times 100$

Answer 1

The correct option is E $\frac{11.1\times 141}{121\times 15}\times 100$

Amount of brnzamide from 15 g benzoyl chloride $=\frac{121}{141}\times 15g$
Actual amount of benzamide formed = 11.1 g

% yield $=\frac{Actualamountofproduct}{Calculatedamountofproduct}\times 100$

$=\frac{11.1\times 100}{(\frac{121}{141}\times 15)}=\frac{11.1\times 141}{121\times 15}\times 100$