[alwaysalone]

The problem: For T= 1500 K and V=1L : CaCO₃(s) :requil:CaO (s) + CO₂(g) with   K= 0.05mol/L  and CO₂(g) ::equil::CO (g) + O₂(g) with X_O2= 0.15 in equilibrium. There is 1 mol of CaCO₃at the begining, then moles CaCO₃in equilibrium.
I try this way:

         CaCO₃(s) =CaO (s) + CO₂(g)    K= 0.05mol/L
initial n       1 mol          0            0
equil n        1-x             x            x

x/1L =K=0.05M = [CO₂] and then I supossed that this quantity is the begining quantity for the other equation:
            CO₂(g) =CO (g) + 1/2O₂(g)
initial n   0.05 M          0           0
 equil n   0.05 -x         x          1/2x

X_o2: 0.15 = 1/2x/(0.05-x+x+0.5x)   x=0.017 M
Now, I've lost myself. ¿What can i do with this x?

[Corribus]

I don't believe you can solve the problem by treating each equilibrium as isolated from the other, because the two equilibria are interconnected.  E.g., at equilibrium, the concentration of CaO and CO₂ are not identical, because CO₂ is consumed in the second reaction. Therefore you cannot solve for x in the first table as though these two species have the same concentration.

I would set up two ICE tables with different variables x and y for changes. The change in CO₂ would then be x-y (gained in first equilibrium and lost in second). Then you can solve for x because you know y = 0.15 mol. A quick solve gives me moles of CaCO₃, CaO and CO₂ at equilibrium equal to 0.721, 0.279, and 0.129, respectively. This does return the equilibrium constant for the first reaction of 0.05, so at least that checks out.