Enzymes convert glucose $(M=180.2)$ to ethanol $(M=46.1)$ according to the equation

\[ \text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_2\text{OH} + 2\text{CO}_2 \]

What is the maximum mass of ethanol that can be made from $15.5$ kg of glucose? 

$ \textbf{(A)}\hspace{.05in}0.256 \text{kg}\qquad\textbf{(B)}\hspace{.05in}0.512 \text{kg} \qquad\textbf{(C)}\hspace{.05in}3.96 \text{kg}\qquad\textbf{(D)}\hspace{.05in}7.93 \text{kg}\qquad$
1. **Calculate the number of moles of glucose:**
   The molar mass of glucose (C\(_6\)H\(_{12}\)O\(_6\)) is given as \(M = 180.2 \, \text{g/mol}\). We have 15.5 kg of glucose, which is equivalent to \(15.5 \times 1000 = 15500 \, \text{g}\).

   \[
   \text{Number of moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{15500 \, \text{g}}{180.2 \, \text{g/mol}} \approx 86 \, \text{mol}
   \]

2. **Determine the number of moles of ethanol produced:**
   According to the balanced chemical equation:

   \[
   \text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2
   \]

   1 mole of glucose produces 2 moles of ethanol. Therefore, 86 moles of glucose will produce:

   \[
   86 \, \text{mol} \times 2 = 172 \, \text{mol of ethanol}
   \]

3. **Calculate the mass of ethanol produced:**
   The molar mass of ethanol (C\(_2\)H\(_5\)OH) is given as \(M = 46.1 \, \text{g/mol}\).

   \[
   \text{Mass of ethanol} = \text{number of moles of ethanol} \times \text{molar mass of ethanol} = 172 \, \text{mol} \times 46.1 \, \text{g/mol} = 7929.2 \, \text{g}
   \]

4. **Convert the mass of ethanol to kilograms:**

   \[
   \text{Mass of ethanol in kg} = \frac{7929.2 \, \text{g}}{1000} = 7.9292 \, \text{kg}
   \]

   Rounding to the appropriate number of significant figures, we get:

   \[
   7.93 \, \text{kg}
   \]

Conclusion:
\[
\boxed{7.93 \, \text{kg}}
\]