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$20 \mathrm{~mL}$ 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
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The correct answer is:
ethyl alcohol
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.
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.
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