$60{\text{ grams C}}{{\text{H}}_3}{\text{COOH}}$ and $46{\text{ grams }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ react in $5{\text{ L}}$ flask to form $44{\text{ grams C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ at equilibrium on taking $120{\text{ grams C}}{{\text{H}}_3}{\text{COOH}}$ and $46{\text{ grams }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$, ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ formed at equilibrium is
A) $44{\text{ g}}$
B) $20.33{\text{ g}}$
C) $22{\text{ g}}$
D) $58.66{\text{ g}}$
Answer
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Hint: First calculate the equilibrium constant using the amount of the reactants and products given. Then in the second case calculate the concentration of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ formed using the value of equilibrium constant.
Formulae Used:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
Complete step by step answer:We know that the molar mass of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ is $60{\text{ g mo}}{{\text{l}}^{ - 1}}$, the molar mass of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ is $46{\text{ g mo}}{{\text{l}}^{ - 1}}$ and the molar mass of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ is $88{\text{ g mo}}{{\text{l}}^{ - 1}}$.
Calculate the initial number of moles of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{60 g}}}}{{{\text{60 g mo}}{{\text{l}}^{ - 1}}}} = 1{\text{ mol}}$
Calculate the initial molar concentration of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{1 mol}}}}{{{\text{5 L}}}} = 0.2{\text{ mol }}{{\text{L}}^{ - 1}}$
Calculate the initial number of moles of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{46 g}}}}{{{\text{46 g mo}}{{\text{l}}^{ - 1}}}} = 1{\text{ mol}}$
Calculate the initial molar concentration of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{1 mol}}}}{{{\text{5 L}}}} = 0.2{\text{ mol }}{{\text{L}}^{ - 1}}$
Calculate the number of moles of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ at equilibrium using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5} = \dfrac{{{\text{44 g}}}}{{{\text{88 g mo}}{{\text{l}}^{ - 1}}}} = 0.5{\text{ mol}}$
Calculate the molar concentration of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ at equilibrium using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5} = \dfrac{{{\text{0}}{\text{.5 mol}}}}{{{\text{5 L}}}} = 0.1{\text{ mol }}{{\text{L}}^{ - 1}}$
The reaction of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ with ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ to produce ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ is as follows:
${\text{C}}{{\text{H}}_3}{\text{COOH}} + {{\text{C}}_2}{{\text{H}}_5}{\text{OH}} \to {\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5} + {{\text{H}}_2}{\text{O}}$
Initially there are $0.2{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $0.2{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
From the reaction, we can see that one mole of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ reacts with one mole of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ to produce one mole of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$.
At equilibrium $0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ are produced. Thus, at equilibrium to produce $0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$we require $\left( {0.2 - 0.1} \right){\text{ mol }}{{\text{L}}^{ - 1}} = 0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $\left( {0.2 - 0.1} \right){\text{ mol }}{{\text{L}}^{ - 1}} = 0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
Thus, the equation for equilibrium constant is as follows:
$K = \dfrac{{{\text{[C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}{\text{][}}{{\text{H}}_2}{\text{O]}}}}{{{\text{[C}}{{\text{H}}_3}{\text{COOH][}}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}]}}$
$K = \dfrac{{0.1 \times 0.1}}{{0.1 \times 0.1}}$
$K = 1$
In the second case,
Calculate the initial number of moles of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{120 g}}}}{{{\text{60 g mo}}{{\text{l}}^{ - 1}}}} = 2{\text{ mol}}$
Calculate the initial molar concentration of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{2 mol}}}}{{{\text{5 L}}}} = 0.4{\text{ mol }}{{\text{L}}^{ - 1}}$
Calculate the initial number of moles of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{46 g}}}}{{{\text{46 g mo}}{{\text{l}}^{ - 1}}}} = 1{\text{ mol}}$
Calculate the initial molar concentration of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{1 mol}}}}{{{\text{5 L}}}} = 0.2{\text{ mol }}{{\text{L}}^{ - 1}}$
Initially there are $0.4{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $0.2{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
From the reaction, we can see that one mole of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ reacts with one mole of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ to produce one mole of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$.
Consider that at equilibrium $x{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ are produced. Thus, at equilibrium to produce $x{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$we require $\left( {0.4 - x} \right){\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $\left( {0.2 - x} \right){\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
Thus, the equation for equilibrium constant is as follows:
$K = \dfrac{{{\text{[C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}{\text{][}}{{\text{H}}_2}{\text{O]}}}}{{{\text{[C}}{{\text{H}}_3}{\text{COOH][}}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}]}}$
$K = \dfrac{{{x^2}}}{{\left( {0.4 - x} \right)\left( {0.2 - x} \right)}} = 1$ …… ${\text{[C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}{\text{]}} = {\text{[}}{{\text{H}}_2}{\text{O]}}$
${x^2} = \left( {0.4 - x} \right)\left( {0.2 - x} \right)$
${x^2} = 0.08 + {x^2} - 0.6x$
$0.6x = 0.08$
$x = \dfrac{{0.08}}{{0.6}}$
Thus, the number of moles of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ $ = \dfrac{{0.08}}{{0.6}} \times 5 = 0.6666{\text{ mol}}$
Thus, mass of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ $ = 0.6666 \times 88 = 58.66{\text{ g}}$
Thus, the mass of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ formed at equilibrium is $58.66{\text{ g}}$.
Thus, the correct option is (D) $58.66{\text{ g}}$.
Note:Remember that the equilibrium constant expresses the relationship between the amounts of products and the amounts of reactants that are present at equilibrium in a reversible reaction. In simple words, equilibrium constant is the ratio of the concentration of products to the concentration of reactants.
Formulae Used:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
Complete step by step answer:We know that the molar mass of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ is $60{\text{ g mo}}{{\text{l}}^{ - 1}}$, the molar mass of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ is $46{\text{ g mo}}{{\text{l}}^{ - 1}}$ and the molar mass of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ is $88{\text{ g mo}}{{\text{l}}^{ - 1}}$.
Calculate the initial number of moles of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{60 g}}}}{{{\text{60 g mo}}{{\text{l}}^{ - 1}}}} = 1{\text{ mol}}$
Calculate the initial molar concentration of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{1 mol}}}}{{{\text{5 L}}}} = 0.2{\text{ mol }}{{\text{L}}^{ - 1}}$
Calculate the initial number of moles of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{46 g}}}}{{{\text{46 g mo}}{{\text{l}}^{ - 1}}}} = 1{\text{ mol}}$
Calculate the initial molar concentration of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{1 mol}}}}{{{\text{5 L}}}} = 0.2{\text{ mol }}{{\text{L}}^{ - 1}}$
Calculate the number of moles of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ at equilibrium using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5} = \dfrac{{{\text{44 g}}}}{{{\text{88 g mo}}{{\text{l}}^{ - 1}}}} = 0.5{\text{ mol}}$
Calculate the molar concentration of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ at equilibrium using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5} = \dfrac{{{\text{0}}{\text{.5 mol}}}}{{{\text{5 L}}}} = 0.1{\text{ mol }}{{\text{L}}^{ - 1}}$
The reaction of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ with ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ to produce ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ is as follows:
${\text{C}}{{\text{H}}_3}{\text{COOH}} + {{\text{C}}_2}{{\text{H}}_5}{\text{OH}} \to {\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5} + {{\text{H}}_2}{\text{O}}$
Initially there are $0.2{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $0.2{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
From the reaction, we can see that one mole of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ reacts with one mole of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ to produce one mole of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$.
At equilibrium $0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ are produced. Thus, at equilibrium to produce $0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$we require $\left( {0.2 - 0.1} \right){\text{ mol }}{{\text{L}}^{ - 1}} = 0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $\left( {0.2 - 0.1} \right){\text{ mol }}{{\text{L}}^{ - 1}} = 0.1{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
Thus, the equation for equilibrium constant is as follows:
$K = \dfrac{{{\text{[C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}{\text{][}}{{\text{H}}_2}{\text{O]}}}}{{{\text{[C}}{{\text{H}}_3}{\text{COOH][}}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}]}}$
$K = \dfrac{{0.1 \times 0.1}}{{0.1 \times 0.1}}$
$K = 1$
In the second case,
Calculate the initial number of moles of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{120 g}}}}{{{\text{60 g mo}}{{\text{l}}^{ - 1}}}} = 2{\text{ mol}}$
Calculate the initial molar concentration of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of C}}{{\text{H}}_3}{\text{COOH}} = \dfrac{{{\text{2 mol}}}}{{{\text{5 L}}}} = 0.4{\text{ mol }}{{\text{L}}^{ - 1}}$
Calculate the initial number of moles of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
${\text{Number of moles of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{46 g}}}}{{{\text{46 g mo}}{{\text{l}}^{ - 1}}}} = 1{\text{ mol}}$
Calculate the initial molar concentration of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume of solution}}}}$
${\text{Molarity of }}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}} = \dfrac{{{\text{1 mol}}}}{{{\text{5 L}}}} = 0.2{\text{ mol }}{{\text{L}}^{ - 1}}$
Initially there are $0.4{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $0.2{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
From the reaction, we can see that one mole of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ reacts with one mole of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$ to produce one mole of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$.
Consider that at equilibrium $x{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ are produced. Thus, at equilibrium to produce $x{\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$we require $\left( {0.4 - x} \right){\text{ mol }}{{\text{L}}^{ - 1}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and $\left( {0.2 - x} \right){\text{ mol }}{{\text{L}}^{ - 1}}$ of ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}$.
Thus, the equation for equilibrium constant is as follows:
$K = \dfrac{{{\text{[C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}{\text{][}}{{\text{H}}_2}{\text{O]}}}}{{{\text{[C}}{{\text{H}}_3}{\text{COOH][}}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}]}}$
$K = \dfrac{{{x^2}}}{{\left( {0.4 - x} \right)\left( {0.2 - x} \right)}} = 1$ …… ${\text{[C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}{\text{]}} = {\text{[}}{{\text{H}}_2}{\text{O]}}$
${x^2} = \left( {0.4 - x} \right)\left( {0.2 - x} \right)$
${x^2} = 0.08 + {x^2} - 0.6x$
$0.6x = 0.08$
$x = \dfrac{{0.08}}{{0.6}}$
Thus, the number of moles of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ $ = \dfrac{{0.08}}{{0.6}} \times 5 = 0.6666{\text{ mol}}$
Thus, mass of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ $ = 0.6666 \times 88 = 58.66{\text{ g}}$
Thus, the mass of ${\text{C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}$ formed at equilibrium is $58.66{\text{ g}}$.
Thus, the correct option is (D) $58.66{\text{ g}}$.
Note:Remember that the equilibrium constant expresses the relationship between the amounts of products and the amounts of reactants that are present at equilibrium in a reversible reaction. In simple words, equilibrium constant is the ratio of the concentration of products to the concentration of reactants.
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