What concentration of `CO_(2)` be in equilibrium with `0.025 M` CO at `120^(@)C` for the reaction
`FeO(s)+CO(g) hArr Fe(s)+CO_(2)(g)`
if the value of `K_(c)=5.0` ?

Options

A. `0.125 M`
B. `0.0125 M`
C. `1.25 M`
D. `12.5 M`

Explanation

To solve the problem, we need to determine the equilibrium concentration of \( CO_2 \) given the equilibrium concentration of \( CO \) and the equilibrium constant \( K_c \) for the reaction:

\[ FeO(s) + CO(g) \rightleftharpoons Fe(s) + CO_2(g) \]

### Step-by-Step Solution:

1. **Write the expression for the equilibrium constant \( K_c \)**:
The equilibrium constant \( K_c \) for the reaction is given by the formula:
\[ K_c = \frac{[CO_2]}{[CO]} \]
where \([CO_2]\) is the concentration of carbon dioxide and \([CO]\) is the concentration of carbon monoxide.

2. **Substitute the known values**:
We know that:
- \( K_c = 5.0 \)
- \([CO] = 0.025 \, M\)

Plugging these values into the equation gives:
\[ 5.0 = \frac{[CO_2]}{0.025} \]

3. **Rearrange the equation to solve for \([CO_2]\)**:
To find \([CO_2]\), we rearrange the equation:
\[ [CO_2] = K_c \times [CO] \]
Substituting the values we have:
\[ [CO_2] = 5.0 \times 0.025 \]

4. **Calculate \([CO_2]\)**:
Now, performing the multiplication:
\[ [CO_2] = 5.0 \times 0.025 = 0.125 \, M \]

5. **Conclusion**:
The equilibrium concentration of \( CO_2 \) is:
\[ [CO_2] = 0.125 \, M \]

### Final Answer:
The concentration of \( CO_2 \) in equilibrium with \( 0.025 \, M \) \( CO \) at \( 120^\circ C \) is \( 0.125 \, M \).