How many grams of each substance will be present at equilibrium in the reaction of ethanol and ethanoic acid?
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... :/