How many grams of each substance will be present at equilibrium in the reaction of ethanol and ethanoic acid? In the following reaction, $K_c=4.0$:
  $$\ce{C2H5OH + CH3COOH &lt;=&gt; CH3COOC2H5 + H2O}$$
  
  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.

$$\begin{align} 
[\ce{C2H5OH}] 
  &amp;= \pu{17.2 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} &amp;
  &amp;= \pu{0.373 mol} \\
[\ce{CH3COOH}] 
  &amp;= \pu{23.8 g} \times \frac{\pu{1mol}}{\pu{60.1 g}} &amp;
  &amp;= \pu{0.396mol} \\
[\ce{CH3COOC2H5}] 
  &amp;= \pu{48.6 g} \times \frac{\pu{1 mol}}{\pu{88.1 g}} &amp; 
  &amp;= \pu{0.552mol} \\
[\ce{H2O}] 
  &amp;= \pu{71.2 g} \times \frac{\pu{1 mol}}{\pu{18.0 g}} &amp;
  &amp;= \pu{3.96 mol} \\
\end{align}$$

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?
Knowing that $K_c=4.0$, equilibrium is achieved when $Q_c = K_c$.

Let's set up an ICE table and write down our equilibrium expression like you did:

Reactants:

$$\ce{C2H5OH + CH3COOH &lt;=&gt; CH3COOC2H5 + H2O}$$

\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} &amp; \pu{0.373 mol} &amp; \pu{0.396 mol} &amp; \pu{0.552 mol} &amp; \pu{3.96 mol}\\
\hline
\text{Change conc.} &amp; -x &amp; -x &amp; +x &amp; +x\\
\hline
\text{End conc.} &amp; \pu{0.373 mol} - x &amp; \pu{0.396 mol} - x &amp; \pu{0.552 mol} + x &amp; \pu{3.96 mol} + x \\
\hline
\end{array}

Assuming that our reaction is at equilibrium:

$$ K_c = \ce{\frac{[CH3COOC2H5][H2O]}{[C2H5OH][CH3COOH]}}$$

We plug in our values for our change in equilibrium:

$$ K_c = 4.0 = {\frac{[0.552 + x][3.96 + x]}{[0.373 - x][0.396 - x]}}$$

Solving for $x$, which is the change in concentration:

\begin{align}
4 &amp;= \frac{x^2+4.512 x+2.18592}{x^2-0.769 x+0.147708}\\
0 &amp;= 3x^2-7.588 x-1.59509 
\end{align}

Using the quadratic equation, or thankfully polysolv on modern calculators, we get two values for $x$:

\begin{align}
x_1 &amp;= 2.72449 \\
x_2 &amp;= -0.195154
\end{align}

The negative zero is chosen, since the reaction will evolve to the left (being $Q_c&gt;K_c$), thus increasing the concentration of the reactants. This will only be accounted for, if one takes the negative zero. If one chooses the other zero, one will go to the wrong opposite direction (towards a diminution of reactants concentration). In the particular case of the present exercise, one would even result in a negative concentration. Therefore:

\begin{align}
\ce{[CH3COOH]} 
  &amp;= \pu{0.373 mol} - (\pu{-0.195154 mol}) &amp;
  &amp;= 0.568154\\
\ce{[C2H5OH]} 
  &amp;= \pu{0.396 mol} - (\pu{-0.195154 mol}) &amp;
  &amp;= 0.591154\\
\ce{[CH3COOC2H5]} 
  &amp;= \pu{0.552 mol} + (\pu{-0.195154 mol}) &amp;
  &amp;= 0.356846\\
\ce{[H2O]} 
  &amp;= \pu{3.96 mol} + (\pu{-0.195154 mol}) &amp;
  &amp;= 3.764846
\end{align}

Therefore:

\begin{align}
\pu{0.568154 mol} \ \ce{CH3COOH} \times \frac{\pu{60.1 g}}{\pu{1 mol} \ \ce{CH3OOH}} &amp;= \pu{34.1 g}\\
\pu{0.591154 mol} \ \ce{C2H5OH} \times \frac{\pu{46.1 g}}{\pu{1 mol} \ \ce{C2H5OH}} &amp;= \pu{27.3 g}\\
\pu{0.356846 mol} \ \ce{CH3COOC2H5} \times \frac{\pu{88.1 g}}{\pu{1 mol} \ \ce{CH3COOC2H5}} &amp;= \pu{31.4 g}\\
\pu{3.764846 mol} \ \ce{H2O} \times \frac{\pu{18.0 g}}{\pu{1 mol} \ \ce{H2O}} &amp;= \pu{67.8 g}
\end{align}

Initial product sum is $\pu{160.8 g}$.

Equilibrium product sum is $\pu{160.6 g}$. $\pu{0.2 g}$ is likely due to truncation error in calculations. 

This problem is really cruel... whoever assigned it... :/