Question

When 0.100 mol of CaCO3(s) and 0.100 mol of CaO(s) are placed in an evacuated, sealed 10.0−L container and heated to 385 K, PCO2 = 0.220 atm after equilibrium is established: CaCO3(s) ⇌ CaO(s) + CO2(g) An additional 0.290 atm of CO2(g) is pumped in. What is the total mass (in g) of CaCO3 after equilibrium is reestablished?

Answer 1

CaCO₃(s) <-----> CaO(s) + CO₂(g)

pCO₂(g) = 0.220 atm.

We have, ideal gas equation

PV = nRT

Rearranging,

n = PV/RT = (0.220 x 10)/(0.08206 x 385) = 0.0696 moles

Thus, 0.0696 moles of CaCO3 has been converted to CaO before pumping 0.290 atm CO₂(g).

After pumping 0.290 atm CO₂(g), the equilibrium pressure of CO₂ is 0.220 atm. Therefore, the additional CO₂ must have converted to CaCO₃.

Now,

n = PV/RT = 0.290 x 10/(0.08206 x 385) = 0.0917 moles

Therefore, total CaCO₃ after the equilibrium is reestablished = 0.100 - 0.0696 + 0.0917 = 0.1221 moles

Molar mass of CaCO₃ = 100.0869 g/mol

Therefore, mass of CaCO₃ after the equilibrium is reestablished = 0.1221 moles x 100.0869 g/mol = 12.22 g

Mass of CaCO₃ after the equilibrium is reestablished = 12.22 g