For a reaction `[CaCO_3(s) -> CaO(s) + CO (g)]` `(K_c = 0.5 moll^-1)`, the maximum moles of CO (g) formed at equilibrium in 5 L container is

Options

A. 5
B. 2.5
C. 1.24
D. 10

Answer

To solve the problem, we need to determine the maximum moles of CO (carbon monoxide) formed at equilibrium in a 5 L container for the reaction:

\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}(g) \]

Given that the equilibrium constant \( K_c \) for the reaction is \( 0.5 \, \text{mol}^{-1} \).

### Step-by-Step Solution:

1. **Identify the Reaction and Components**:
- The reaction involves solid calcium carbonate decomposing into solid calcium oxide and gaseous carbon monoxide.
- The equilibrium constant \( K_c \) is only applicable to the gaseous products.

2. **Write the Expression for \( K_c \)**:
- The expression for the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = \frac{[\text{CO}]}{1} \]
Since solids do not appear in the equilibrium expression, we only consider the concentration of CO.

3. **Substitute the Given \( K_c \)**:
- We know that \( K_c = 0.5 \, \text{mol}^{-1} \).
- Therefore, we can write:
\[ 0.5 = [\text{CO}] \]

4. **Relate Concentration to Moles**:
- The concentration of CO is defined as:
\[ [\text{CO}] = \frac{\text{Number of moles of CO}}{\text{Volume of the container in liters}} \]
- In this case, the volume of the container is 5 L.

5. **Set Up the Equation**:
- Let \( n \) be the number of moles of CO. Then:
\[ 0.5 = \frac{n}{5} \]

6. **Solve for \( n \)**:
- Rearranging the equation gives:
\[ n = 0.5 \times 5 \]
- Calculating this:
\[ n = 2.5 \, \text{moles} \]

### Final Answer:
The maximum moles of CO (g) formed at equilibrium in a 5 L container is **2.5 moles**.