1. Write the balanced equilibrium

$2\,NOCl(g) \rightleftharpoons 2\,NO(g) + Cl_2(g)$

2. Initial concentrations

Since $2.0$ moles of $NOCl$ are placed in a $1$ L flask,

$[NOCl]_0 = 2.0\,\text{M}$

Initially, no products are present:

$[NO]_0 = 0, \qquad [Cl_2]_0 = 0$

3. Use the equilibrium information

Given:

$[NO]_{eq} = 0.4\,\text{M}$

From the stoichiometry

$2\,NOCl \rightarrow 2\,NO + Cl_2$

The change in $NO$ is $+0.4$ M, so the change in $NOCl$ is $-0.4$ M, and the change in $Cl_2$ is $+0.2$ M.

Thus,

$[NOCl]_{eq} = 2.0 - 0.4 = 1.6\,\text{M}$ $[NO]_{eq} = 0.4\,\text{M}$ $[Cl_2]_{eq} = 0.2\,\text{M}$

4. Expression for equilibrium constant

For the reaction,

$K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2}$

Substitute the equilibrium concentrations:

$K_c = \frac{(0.4)^2(0.2)}{(1.6)^2}$

5. Calculate

$(0.4)^2 = 0.16$ $0.16 \times 0.2 = 0.032$ $(1.6)^2 = 2.56$

So,

$K_c = \frac{0.032}{2.56} = 0.0125$

Now write in the form $\_\_\_ \times 10^{-4}$:

$0.0125 = 125 \times 10^{-4}$

6. Final integer answer

$\boxed{125}$