## Chemistry (7th Edition)

No, the amount of remaining ethanol will be very insignificant. Since this reaction has a very large equilibrium constant $(1.2 \times 10^{82})$, the reactants are negligible at the equilibrium.
Analyzing mathematically the expression: $K_c = \frac{[Products]}{[Reactants]}$, it is possible to predict that: as the products concentration decrease, the $K_c$ gets smaller, and as the reactants concentration decrease, the $K_c$ gets larger. Since the $K_c$ has a very large value in this case $(1.2 \times 10^{82})$, there is a very small reactants concentration. Knowing that Ethanol is a reactant, we can affirm that its concentration is going to be very small too.