Chemistry Help Please!!?

Question

When 0.100 mol CaCO3(s) and 0.100 mol CaO(s) are placed
in an evacuated sealed 11.5 L container and heated to 385 K, P(CO2) = 0.220 atm after equilibrium is established.

CaCO3(s) <~> CaO(s) + CO2(g)

An additional 0.225 atm CO2(g) is then pumped into the container. What is the total mass (in g) of CaCO3 after equilibrium is reestablished?
_____ g

Answer 1

First, I do not believe that the reaction could take place at temperature as low as 385K.
Now, suppose it does.
CaCO3(s) <==> CaO(s) + CO2(g)
reaction constant K = [CO2(g)] = 0.220 atm.
Use ideal gas fomula, we have
n = PV/RT = 0.220*11.5/(0.08206*385) = 0.0801 (mol)
this is the amount of CaCO3 which has been converted to CaO before pumping-in additional 0.225 atm CO2(g).
After pumping-in additional 0.225 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm (do you see why?). Therefore all this additional CO2 would completely convert to CaCO3:
n = PV/RT = 0.225*11.5/(0.08206*385) = 0.0819 (mol)
Hence the total CaCO3 after equilibrium is reestablished is:
0.100 - 0.0801 + 0.0819 (mol) = 0.102 mol
which is 10.2 g of CaCO3.