4 mole of `Sl_(2)Cl_(4)(g)` is introduced innto a `10L` vessel. The following equilibrium was established `S_(2)Cl_(4)(g)hArr2SCl_(2)(g)`
at equilibrium `0.2` mol of `S_(2)Cl_(4)` was present in the vessel. The value of equilibrium constant is.

Explanation

To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction:

\[ S_2Cl_4(g) \rightleftharpoons 2SCl_2(g) \]

### Step-by-Step Solution:

1. **Identify Initial Moles and Volume**:
- Initial moles of \( S_2Cl_4 \) = 4 moles
- Volume of the vessel = 10 L

2. **Determine Moles at Equilibrium**:
- At equilibrium, it is given that 0.2 moles of \( S_2Cl_4 \) are present.

3. **Calculate Change in Moles**:
- Change in moles of \( S_2Cl_4 \):
\[ \text{Change} = \text{Initial moles} - \text{Equilibrium moles} = 4 - 0.2 = 3.8 \text{ moles} \]
- According to the stoichiometry of the reaction, for every 1 mole of \( S_2Cl_4 \) that reacts, 2 moles of \( SCl_2 \) are produced. Thus:
\[ \text{Moles of } SCl_2 = 2 \times \text{Change in moles of } S_2Cl_4 = 2 \times 3.8 = 7.6 \text{ moles} \]

4. **Calculate Concentrations**:
- Concentration of \( S_2Cl_4 \):
\[ [S_2Cl_4] = \frac{0.2 \text{ moles}}{10 \text{ L}} = 0.02 \text{ M} \]
- Concentration of \( SCl_2 \):
\[ [SCl_2] = \frac{7.6 \text{ moles}}{10 \text{ L}} = 0.76 \text{ M} \]

5. **Write the Expression for \( K_c \)**:
- The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[SCl_2]^2}{[S_2Cl_4]} \]

6. **Substitute the Concentrations into the Expression**:
- Substitute the values we calculated:
\[ K_c = \frac{(0.76)^2}{0.02} \]

7. **Calculate \( K_c \)**:
- Calculate \( (0.76)^2 \):
\[ (0.76)^2 = 0.5776 \]
- Now calculate \( K_c \):
\[ K_c = \frac{0.5776}{0.02} = 28.88 \]

### Final Answer:
The value of the equilibrium constant \( K_c \) is approximately **28.88 M**.