At a given temperature, Kc is 4 for the reaction ${{\mathrm{H}}_2}_{\left(\mathrm{g}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}\iff {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{CO}}_{\left(\mathrm{g}\right)}$Initially 0.6 moles each of ${\mathrm{H}}_2 \mathrm{and} {\mathrm{CO}}_2$are taken in 1lit flask. The equilibrium concentration of ${\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)} \mathrm{is}$

The equilibrium reaction is

${{\mathrm{H}}_2}_{\left(\mathrm{g}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}\iff {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{CO}}_{\left(\mathrm{g}\right)}$

Initial(moles):    0.6     0.6       0    0

Change(moles):  -x      -x         +x   +x

Equilib(moles): 0.6 - x   0.6 -x  x   x    From equilibrium constant expression,

${\mathrm{K}}_{\mathrm{C}}=\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[\mathrm{CO}\right]}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{CO}}_2\right]}=\frac{\left[\mathrm{Conc}.\mathrm{of} \mathrm{products}\right]}{\left[\mathrm{Conc}.\mathrm{of} \mathrm{products}\right]}$

$4=\frac{\mathrm{x}\times \mathrm{x}}{(0.6-\mathrm{x})(0.6-\mathrm{x})}$

${\left(2\right)}^2={\left[\frac{\mathrm{x}}{(0.6-\mathrm{x})}\right]}^2$

Apply square root on both sides to the above equation.

$\frac{2}{1}\times \frac{x}{(0.6-x)}$

$\Rightarrow \mathrm{x}=1.2-2\mathrm{x}$

$\therefore \mathrm{x}=0.4$

Therefore, concentration of ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \mathrm{is}$

Molarity of ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=\frac{\mathrm{n}}{\mathrm{V}}$

Molarity of ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=\frac{0.4 \mathrm{mole}}{\mathrm{IL}}=0.4\mathrm{M}$