Q1 / 180 · COMEDK 2015 · Chemistry

$20 \mathrm{~mL}$ of acetic acid reacts with $20 \mathrm{~mL}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{~g} / \mathrm{mL}$ and $0.7 \mathrm{~g} / \mathrm{mL}$ respectively. The limiting reagent in this reaction is

  • A
    acetic acid
  • Correct answer
    B
    ethyl alcohol
  • C
    acetic acid and ethyl alcohol
  • D
    ester

Solution

Mass of $20 \mathrm{~mL}$ of acetic acid $=20 \times 1=20 \mathrm{~g}$ Mass of $20 \mathrm{~mL}$ of ethyl alcohol $=20 \times 0.7=14 \mathrm{~g}$ For the reaction, $$ \underset{60 \mathrm{~g}}{\mathrm{CH}_{3} \mathrm{COOH}}+\underset{46 \mathrm{~g}}{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}} \underset{-\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} $$ $60 \mathrm{~g}$ of acetic acid requires $=46 \mathrm{~g}$ of ethyl alcohol $20 \mathrm{~g}$ of acetic acid requires $=\frac{46}{60} \times 20=15.3 \mathrm{~g}$ of ethyl alcohol Therefore, ethyl alcohol is the limiting reagent.

Hence, the correct answer is option: 'B'.

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