When 0.100 mol CaCO3(s) and 0.100 mol CaO(s) are placed

Question

in an evacuated sealed 10.0 L container and heated to 385 K, P(CO2) = 0.220 atm after equilibrium is established.
CaCO3(s) <~> CaO(s) + CO2(g)

0.05 atm CO2(g) is then removed from the container. What is the total mass (in g) of CaCO3 after equilibrium is reestablished?
_____ g

Answer 1

I would do this.

...........CaCO3 ==> CaO(s) + CO2
I.........0.1 mol...0.1 mol...0
C...........-x.......+x.......+x
E..........0.1-x....01+x......+x
You know x is 0.220 atm. Use PV = nRT and solve for n = number of mols. x = approx 0.07 mols and (CO2) = approx 0.07/10L = approx 0.007 M.

Do another ICE chart starting the equilibrium conditions. Removing 0.05 atm CO2 means the equilibrium will shift to the right and mols CO2 removed is approx 0.016 (again, you need to redo all this numbers) and the approx 0.016 again comes from PV = nRT.
Plug those values into the new ICE chart and calculate (CO2) from the Kc value shown above. Convert to mols and subtract from the new value of CaCO3. That should give you the mols CaCO3 remaining. Convert mols CaCO3 to grams.
Post your work if you get stuck.