For the reaction that produces ethyl ethanoate from ethanol and ethanoic acid, the equilibrium constant, $K_c$, has a value of $4.0$ at $60^{\circ}\text{C}$. $\text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons \text{CH}_3\text{CO}_2\text{C}_2\text{H}_5 + \text{H}_2\text{O}$ If $1.0\,\text{mol}$ of ethanol and $1.0\,\text{mol}$ of ethanoic acid are left to attain equilibrium at $60^{\circ}\text{C}$, how many moles of ethyl ethanoate are produced?
- A$\frac{1}{3}$
- B$\frac{2}{3}$
- C$\frac{1}{4}$
- D$\frac{3}{4}$