One mole ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$ and one mole CO(g) are taken in 10 L flask and heated to 725 K. At equilibrium, 40 % ( by mass) of water reacted with CO(g) as follows 
${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+\mathrm{CO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$
Its K_c value is

Initial moles of water is 1 mol and only 40% of water reacts, this means that reacted moles of water is 0.4 mol and unreacted moles of water is 0.6 mol.

Given reaction is:

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+\mathrm{CO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

According to given reaction, 0.4 moles of water reacts with 0.4 moles of carbon monoxide to form 0.4 moles of hydrogen gas and 0.4 moles of carbon dioxide.

Equimolar concentration of hydrogen gas is 0.04 M $\left(\frac{0.4}{10}=0.04 \mathrm{M}\right)$.

Equimolar concentration of carbon dioxide gas is 0.04 M $\left(\frac{0.4}{10}=0.04 \mathrm{M}\right)$.

Equimolar concentration of water is 0.06 M $\left(\frac{0.6}{10}=0.06 \mathrm{M}\right)$.

Equimolar concentration of carbon dioxide gas is 0.06 M $\left(\frac{0.6}{10}=0.06 \mathrm{M}\right)$.

${\mathrm{k}}_{\mathrm{C}}=\frac{\left[{\mathrm{H}}_2\right]\left[{\mathrm{CO}}_2\right]}{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[\mathrm{CO}\right]}=\frac{0.04\times 0.04}{0.06\times 0.06}=0.444$