In the process called esterification, an alcohol (like ethanol) reacts with a carboxylic acid (like propanoic acid), producing an ester and water:
$$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{COOH}(\ell) \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{COOC}_{2} \mathrm{H}_{5}(\ell)+\mathrm{H}_{2} \mathrm{O}(\ell) $$
$$ \begin{array}{l} \text { ethanol propanoic acid } & \text { ethyl propanoate } & \text { water }
\end{array}$$ Two (2.00) moles each of ethanol and propanoic acid are combined and the system is allowed to come to equilibrium. At equilibrium, are the amounts of ethanol, propanoic acid, ethyl propanoate, and water $1.00 \mathrm{~mol}$ each? Would any ethanol be left? If the system takes 45 minutes to come to equilibrium, how long would it take for $a l l$ of the ethanol and propanoic acid to be converted to ethyl propanoate and water?