In the following reaction, $K_c=4.0$: $$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$
A reaction is allowed to occur in a mixture of $\pu{17.2 g}$ $\ce{C2H5OH}$, $\pu{23.8 g}$ $\ce{CH3COOH}$, $\pu{48.6 g}$ $\ce{CH3COOC2H5}$, and $\pu{71.2 g}$ $\ce{H2O}$
(a) In what direction will a net change occur?
(b) How many grams of each substance will be present at equilibrium?
This is what I did for part (a): $$Q_c = \frac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{C2H5OH}][\ce{CH3COOH}]}$$
First, get the concentration of $\ce{C2H5OH}$, $\ce{CH3COOH}$, $\ce{CH3COOC2H5}$, and $\ce{H2O}$ using the given amount of each substance.
$$\begin{align} [\ce{C2H5OH}] &= \pu{17.2 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} & &= \pu{0.373 mol} \\ [\ce{CH3COOH}] &= \pu{23.8 g} \times \frac{\pu{1mol}}{\pu{60.1 g}} & &= \pu{0.396mol} \\ [\ce{CH3COOC2H5}] &= \pu{48.6 g} \times \frac{\pu{1 mol}}{\pu{88.1 g}} & &= \pu{0.552mol} \\ [\ce{H2O}] &= \pu{71.2 g} \times \frac{\pu{1 mol}}{\pu{18.0 g}} & &= \pu{3.96 mol} \\ \end{align}$$
Assuming the overall volume is $\pu{1 L}$, then $$Q_c = \frac{(\pu{0.552 M})(\pu{3.96 M})}{(\pu{0.373 M})(\pu{0.396 M})} = 14.8 $$
Since $K_c$ is $4.0$, and $Q_c$ is $14.84$, $Q_c$ is larger than $K_c$ and the direction of the reaction is to the left, towards the reactants.
I know how to solve for (a) but I have no idea how to get the amount of each substance at equilibrium for (b). The answer for (b) should be $\pu{68 g}$. I tried doing the ICE chart, but I did not get $\ce{68 g}$. Do I use the ICE chart for this part of the question or which method should I be using?