The esterification reaction is: $$\mathrm{CH}_{3} \mathrm{COOH} + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} + \mathrm{H}_{2} \mathrm{O}$$ For this reaction, the equilibrium constant $K_c$ is $4$. If we start with $1$ mole of each reactant, at equilibrium, $x$ moles of ester are formed. The equilibrium amounts are: - Acid: $(1-x)$ - Alcohol: $(1-x)$ - Ester: $x$ - Water: $x$ $$K_c = \frac{x^2}{(1-x)^2} = 4 \implies \frac{x}{1-x} = 2 \implies x = 0.667 \text{ moles}$$