Ethyl ethanoate undergoes the reaction shown below. $\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5 + \text{H}_2\text{O} \rightleftharpoons \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{CO}_2\text{H}$ \quad $K_c = 0.27$ Ethanol and ethanoic acid were combined in equal amounts and left until equilibrium was reached. When equilibrium was established, the concentrations of ethanoic acid and ethanol were each $0.42\,\text{mol dm}^{-3}$. What is the equilibrium concentration of ethyl ethanoate?
- A$0.22\,\text{mol dm}^{-3}$
- B$0.65\,\text{mol dm}^{-3}$
- C$0.81\,\text{mol dm}^{-3}$
- D$1.54\,\text{mol dm}^{-3}$