Starting with `250 ml` of `Au^(3+)` solution `250 ml` of `Fe^(2+)` solution, the following equilibrium is established `Au^(3+) (aq) +3Fe^(2+)(aq) hArr 3Fe^(3+) (aq) +Au(s)`
At equilibrium the equivalents of `Au^(3+), Fe^(2+), Fe^(3+)` and `Au` are x,y,z and w respectively. Then:

Options

A. `K_(c) = (Z^(3))/(6xy^(3))`
B. `K_(c) = (3Z^(3))/(2xy^(3))`
C. `K_(c) = (4Z^(3))/(9xy^(3))`
D. `K_(c) = (8Z^(3))/(9 xy^(3))`

Explanation

To solve the problem, we need to analyze the given equilibrium reaction and determine the equilibrium constant \( K_c \) based on the provided information.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The equilibrium reaction is:
\[ \text{Au}^{3+} (aq) + 3 \text{Fe}^{2+} (aq) \rightleftharpoons 3 \text{Fe}^{3+} (aq) + \text{Au} (s) \]

2. **Determine Initial Conditions**:
We start with:
- Volume of \( \text{Au}^{3+} \) solution = 250 mL
- Volume of \( \text{Fe}^{2+} \) solution = 250 mL
- Total volume = 250 mL + 250 mL = 500 mL = 0.5 L

3. **Define Equivalents at Equilibrium**:
Let:
- Equivalents of \( \text{Au}^{3+} \) at equilibrium = \( x \)
- Equivalents of \( \text{Fe}^{2+} \) at equilibrium = \( y \)
- Equivalents of \( \text{Fe}^{3+} \) at equilibrium = \( z \)
- Equivalents of \( \text{Au} \) at equilibrium = \( w \)

4. **Write the Expression for the Equilibrium Constant \( K_c \)**:
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[\text{Fe}^{3+}]^3}{[\text{Au}^{3+}][\text{Fe}^{2+}]^3} \]

5. **Substitute Concentrations**:
The concentrations can be expressed in terms of equivalents and total volume:
- \([\text{Fe}^{3+}] = \frac{z}{0.5}\)
- \([\text{Au}^{3+}] = \frac{x}{0.5}\)
- \([\text{Fe}^{2+}] = \frac{y}{0.5}\)

Thus, substituting these into the \( K_c \) expression gives:
\[ K_c = \frac{\left(\frac{z}{0.5}\right)^3}{\left(\frac{x}{0.5}\right)\left(\frac{y}{0.5}\right)^3} \]

6. **Simplify the Expression**:
Simplifying the expression:
\[ K_c = \frac{z^3}{x \cdot y^3} \cdot \left(\frac{1}{0.5^3}\right) \cdot \left(\frac{1}{0.5}\right) = \frac{z^3 \cdot 8}{x \cdot y^3} \]

7. **Account for Stoichiometry**:
Since \( 3 \text{Fe}^{2+} \) produces \( 3 \text{Fe}^{3+} \) and \( 1 \text{Au}^{3+} \) produces \( 1 \text{Au} \), we can relate \( x \), \( y \), and \( z \) based on stoichiometry:
- From the reaction, \( z = \frac{y}{3} \) and \( x = \frac{y}{3} \).

8. **Final Expression for \( K_c \)**:
Substitute \( z \) and \( x \) in terms of \( y \):
\[ K_c = \frac{\left(\frac{y}{3}\right)^3 \cdot 8}{\left(\frac{y}{3}\right) \cdot y^3} = \frac{\frac{y^3}{27} \cdot 8}{\frac{y^4}{3}} = \frac{8}{27} \cdot \frac{3}{y} = \frac{24}{27y} \]

### Conclusion:
The equilibrium constant \( K_c \) can be expressed in terms of the equivalents at equilibrium. The final answer is:
\[ K_c = \frac{24}{27y} \]