Question

Consider the above reaction where $6.1g$ of benzoic acid is used to get $7.8g$ of $m$ -bromo benzoic acid. The percentage yield of the product is ____.
(Round off to the Nearest integer)
[Given : Atomic masses : $C=12.0u,H:1.0u,O:16.0u,Br=80.0u]$

Answer 1

Answer: 78

Moles of Benzoic acid $=\frac{6.1}{122}$

= moles of $m$ -bromobenzoic acid

So, weight of $m$ -bromobenzoic acid

$=\frac{6.1}{122}\times 201gm$

$=10.05gm$

$\%$ yield $=\frac{\text{ Actual weight }}{\text{ Theoretical weight }}\times 100$

$=\frac{7.8}{10.05}\times 100$

$=77.61\%$