23 g of ethanol (C2H5OH) on reaction with ethanoic acid (CH3COOH) forms 44 g of ethyl ethanoate (CH3COOC2H5) and some weight of water. The percentage yield of the reaction is

Question

23 g of ethanol (C2H5OH) on reaction with ethanoic acid (CH3COOH) forms 44 g of ethyl ethanoate (CH3COOC2H5) and some weight of water. The percentage yield of the reaction is (A) 50% (B) 100% (C) 25% (D) 12.5%

Tardigrade Admin · Accepted Answer

C2H5OH + CH3COOH → CH3COOC2H5 + H2O 1 mole C2H5OH gives 1 moles CH3COOC2H5 44g C2H5OH gives 88g CH3COOC2H5 23g C2H5OH gives 44g CH3COOC2H5 Yield (44/44)×100

Q. $23 g$ of ethanol $(C_{2} H_{5} O H)$ on reaction with ethanoic acid $(C H_{3} COO H)$ forms $44 g$ of ethyl ethanoate $(C H_{3} COO C_{2} H_{5})$ and some weight of water. The percentage yield of the reaction is

$50$

$100$

$25$

$12.5$

$C_{2} H_{5} O H + C H_{3} COO H \to C H_{3} COO C_{2} H_{5} + H_{2} O$
$1$ mole $C_{2} H_{5} O H$ gives $1$ moles $C H_{3} COO C_{2} H_{5}$
$44 g C_{2} H_{5} O H$ gives $88 g C H_{3} COO C_{2} H_{5}$
$23 g C_{2} H_{5} O H$ gives $44 g C H_{3} COO C_{2} H_{5}$
Yield $\frac{44}{44} \times 100$

Q. 23g of ethanol (C2​H5​OH) on reaction with ethanoic acid (CH3​COOH) forms 44g of ethyl ethanoate (CH3​COOC2​H5​) and some weight of water. The percentage yield of the reaction is

50

100

25

12.5