${CH}_{3} {COOH}_{(l)} + C_{2} H_{5} {OH}_{(l)} ⇌ {CH}_{3} {COOC}_{2} H_{5 (l)}$ In the
above reaction, one mole of each of acetic acid and alcohol
are heated in the presence of little conc. $H_{2} {SO}_{4}$ on equilibrium being attained

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$\frac{2}{3}$ mole of ethyl acetate is formed

1 mole of ethyl acetate is formed

2 mole of ethyl acetate is formed

$\frac{1}{2}$ mole of ethyl acetate is formed

answer is D.

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Detailed Solution

In presence of little $H_{2} {SO}_{4}$ ,(as catalyst) about 2/3 mole

of each of ${CH}_{3} COOH and C_{2} H_{5} OH$ react to form

2/3 mole of the product at equilibrium.

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