Calculate `K_(c)` for the reaction:
`2H_(2)(g)+S_(2)(g) hArr 2H_(2)S(g)`
if `1.58` mol `H_(2)S, 1.27` mol `H_(2)` and `2.78xx10^(-6)` mol of `S_(2)` are in equilibrium in a flask of capacity `180 L` at `750^(@)C`.

Explanation

To calculate the equilibrium constant \( K_c \) for the reaction

\[ 2H_2(g) + S_2(g) \rightleftharpoons 2H_2S(g) \]

given the equilibrium moles of the reactants and products, follow these steps:

### Step 1: Write down the equilibrium expression for the reaction.
The equilibrium constant \( K_c \) is given by the expression:

\[ K_c = \frac{[H_2S]^2}{[H_2]^2[S_2]} \]

### Step 2: Calculate the concentrations of each species.
Concentration is calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Given:
- Moles of \( H_2S = 1.58 \, \text{mol} \)
- Moles of \( H_2 = 1.27 \, \text{mol} \)
- Moles of \( S_2 = 2.78 \times 10^{-6} \, \text{mol} \)
- Volume = 180 L

Now calculate the concentrations:

1. Concentration of \( H_2S \):

\[ [H_2S] = \frac{1.58 \, \text{mol}}{180 \, \text{L}} = 0.008777 \, \text{mol/L} \]

2. Concentration of \( H_2 \):

\[ [H_2] = \frac{1.27 \, \text{mol}}{180 \, \text{L}} = 0.007056 \, \text{mol/L} \]

3. Concentration of \( S_2 \):

\[ [S_2] = \frac{2.78 \times 10^{-6} \, \text{mol}}{180 \, \text{L}} = 1.544 \times 10^{-8} \, \text{mol/L} \]

### Step 3: Substitute the concentrations into the equilibrium expression.
Now substitute the calculated concentrations into the \( K_c \) expression:

\[ K_c = \frac{(0.008777)^2}{(0.007056)^2 \times (1.544 \times 10^{-8})} \]

### Step 4: Calculate \( K_c \).
1. Calculate \( (0.008777)^2 \):

\[ (0.008777)^2 = 0.0000771 \]

2. Calculate \( (0.007056)^2 \):

\[ (0.007056)^2 = 0.0000498 \]

3. Now substitute these values into the equation:

\[ K_c = \frac{0.0000771}{0.0000498 \times 1.544 \times 10^{-8}} \]

4. Calculate the denominator:

\[ 0.0000498 \times 1.544 \times 10^{-8} = 7.694 \times 10^{-13} \]

5. Finally, calculate \( K_c \):

\[ K_c = \frac{0.0000771}{7.694 \times 10^{-13}} \approx 1.002 \times 10^{8} \, \text{L/mol} \]

### Final Answer:
\[ K_c \approx 1.002 \times 10^{8} \, \text{L/mol} \]

---