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

Question

A 5.9-gram sample of CaCO3 (molar mass = 100.) is placed in an evacuated 2.00L container and heated to 200^oC. Equilibrium is reached when 5.15% of the CaCO3 has decomposed.

1)Write the Kp expression for this reaction and calculate its value at 200K.

2)How many carbon dioxide molecules are there in the container at equilibrium?

3)When the reaction reaches equilibrium what % of the solid in the container is CaO by mass?

4)If the volume of the container is decreased after equilibrium is reached, but kept at 200^oC, will the mass of CaO increase or decrease from its value at equilibrium? Explain.

Answer 1

5.9 g CaCO3 = 5.9/100 moles and
0.059 moles x 0.515 = 0.00304
...........CaCO3(s) ==> CaO(s) + CO2(g)
begin.......0.059.......0.........0
change.....-0.00304....+0.00304..+0.00304
final.....0.05586.......0.00304...0.00304

(a) Kp = pCO2 Use PV = nRT to calculate PCO2.
(b) Remember there are 6.022E23 molecules in 1 mole.
(c) %CaO = (mass CaO/mass CaCO3)*100 = ??
(d) Use Le Chatelier's Principle.