For the reaction
$$
\mathrm{CO}(\mathrm{g}) + \mathrm{H}{2}\mathrm{O}(\mathrm{g}) \Leftrightarrow \mathrm{CO}{2}(\mathrm{s}) + \mathrm{H}_{2}(\mathrm{g}),
$$
the equilibrium constant is given as 5.
If we start with 3 moles of $\mathrm{H}{2}\mathrm{O}$ and 3 moles of $\mathrm{CO}$ in a 1-liter container, we need to find the number of moles of $\mathrm{CO}{2}$ to be added so that the equilibrium concentration of $\mathrm{CO}$ becomes 2 M.
Let's denote the number of moles of $\mathrm{CO}_{2}$ to be added as $x$.
At equilibrium, the moles of each species would be:
- $\mathrm{CO}$: 3 - 2 + 0 = 2 M (as given)
- $\mathrm{H}_{2}\mathrm{O}$: 3 - 2 + 0 = 1 M
- $\mathrm{H}_{2}$: + 2 M (since 3 M - 2 M react)
- $\mathrm{CO}_{2}$: $x$
Based on the law of mass action, the expression for the equilibrium constant $K$ is:
$$
K = \frac{[\mathrm{CO}{2}][\mathrm{H}{2}]}{[\mathrm{CO}][\mathrm{H}_{2}\mathrm{O}]} = 5
$$
Substituting the equilibrium concentrations:
$$
5 = \frac{x \cdot 2}{2 \cdot 1} = x \cdot 1
$$
So,
$$
x = 5
$$
Hence, $\mathrm{CO}_{2}$ should have 5 moles added to achieve the desired equilibrium concentration.
Therefore, the answer is:
C. 5