What concentration of `CO_(2)` be in equilibrium with `2.5xx10^(-2) mol L^(-1)` of CO at `100^(@)C` for the reaction:
`FeO(s)+CO(g) hArr Fe(s)+CO_(2)(g), K_(c)=5.0`

Explanation

To solve the problem, we need to find the concentration of \( CO_2 \) at equilibrium given the concentration of \( CO \) and the equilibrium constant \( K_c \).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction is given as:
\[ \text{FeO(s)} + \text{CO(g)} \rightleftharpoons \text{Fe(s)} + \text{CO}_2(g) \]

2. **Identify Given Information**:
- Concentration of \( CO \) = \( 2.5 \times 10^{-2} \, \text{mol L}^{-1} \)
- Equilibrium constant \( K_c = 5.0 \)

3. **Write the Expression for \( K_c \)**:
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[\text{CO}_2]}{[\text{CO}]} \]
Since \( \text{FeO} \) and \( \text{Fe} \) are solids, their concentrations do not appear in the expression.

4. **Substitute the Known Values into the \( K_c \) Expression**:
Substitute the known concentration of \( CO \) into the \( K_c \) expression:
\[ 5.0 = \frac{[\text{CO}_2]}{2.5 \times 10^{-2}} \]

5. **Solve for the Concentration of \( CO_2 \)**:
Rearranging the equation to solve for \( [\text{CO}_2] \):
\[ [\text{CO}_2] = 5.0 \times (2.5 \times 10^{-2}) \]
Now calculate:
\[ [\text{CO}_2] = 5.0 \times 2.5 \times 10^{-2} = 12.5 \times 10^{-2} \, \text{mol L}^{-1} \]

6. **Final Answer**:
Thus, the concentration of \( CO_2 \) at equilibrium is:
\[ [\text{CO}_2] = 0.125 \, \text{mol L}^{-1} \]