Question

$23\, g$ of ethanol $(C_2H_5OH)$ on reaction with ethanoic acid $(CH_3COOH)$ forms $44\, g$ of ethyl ethanoate $(CH_3COOC_2H_5)$ and some weight of water. The percentage yield of the reaction is

• A. $50%$
• B. $100%$
• C. $25%$
• D. $12.5%$

Answer 1

Answer: (B)

$C_2H_5OH + CH_3COOH \to CH_3COOC_2H_5 + H_2O$
$1$ mole $C_2H_5OH$ gives $1$ moles $CH_3COOC_2H_5$
$44g \,C_2H_5OH$ gives $88g \,CH_3COOC_2H_5$
$23g\, C_2H_5OH$ gives $44g \,CH_3COOC_2H_5$
Yield $\frac{44}{44}\times100$