[Win,odd Dhamnekar]

 Consider the reaction CO(g) + H₂O(g)    CO₂(g) + H_2(g)

At some temperature, the equilibrium mixture consists of 0.30 mol of CO₂, 0.30 mol of H₂, 0.10 mol of CO and 0.20 mol of H₂O in a 1.00-L vessel.

So, the equilibrium constant K_c= [itex]\frac{[0.3]^2}{(0.10)(0.20)}=4.5[/itex]

Now, how much H₂ must be removed at constant temperature and volume in order to increase the concentration of CO₂ to 0.40 mol/L?

 The answer is because the stoichiometry of the reaction indicates that 1.00 mol of CO₂ is produced from 1.00 mol of CO, then in order to increase the number of moles of CO₂ from 0.30 to 0.40 requires 0.10 mol of CO. Therefore, all of the CO in this system must be consumed by the reaction, and to accomplish this , all of the H₂ (g) would have to be removed.

I want to know, in that case, how would second  K_c look like? Is it [itex]\frac{[0.4]}{[0.4]^2}=2.5[/itex]

Will the chemical reaction change to CO + O₂  CO₂

[Aldebaran]

In order to have an equilibrium constant there needs to be an equilibrium. If you keep taking away one of the products until there is no more reactant you no longer have an equilibrium .

[Hunter2]

Additionally: if CO2 is increased also H2 has to be increased, because both on product side. The exercise  is so  not solouble.