What would the % yield of ester be if the reaction were simply allowed to equilibrate (i.e. no products are removed)?

C₃H₇OH+C₂H₅COOH ----> C₆H₁₂O₂+H₂O

- Percent yield of ester was 71.6%
- weight of product= 20.7868g
- 19.0mL (0.25mol) of n- propanol
- 18.5 mL (0.25 mol) of propionic acid
- At room temperature, the equilibrium constant, Keq is approximately 3.