Question

Calcium carbonate decomposes at high temperatures to give calcium oxide and carbon dioxide as shown below. CaCO3(s) CaO(s) + CO2(g) The KP for this reaction is 1.16 at 800°C. A 5.00 L vessel containing 10.0 g of CaCO3(s) was evacuated to remove the air, sealed, and then heated to 800°C. Ignoring the volume occupied by the solid, what will be the overall mass percent of carbon in the solid once equilibrium is reached?

A)           5.36% carbon by mass

B)           5.56% carbon by mass

C)           5.76% carbon by mass

D)           5.96% carbon by mass

E)            None of the above

Answer 1

CaCO₃(s) ------> CaO(s) + CO₂(g)

K_p = P_CO2

Molar mass of CaCO₃ = 100 g

Thus, moles of CaCO₃ present in 10 g of it = mass/molar mass = 10/100 = 0.1

moles of CO₂ formed = (P*V)/(R*T) = (1.16*5)/(0.0821*1073) = 0.066

Thus, At eqb., moles of CaCO₃ left unreacted = initial moles of CaCO₃ - moles of CO₂ formed = 0.034

mass of CaCO₃ left unreacted = moles*molar mass = 3.4 g

Thus, mass of solid carbon = moles of unreacted CaCO₃ *molar mass of C = 0.034*12 = 0.408

moles of CaO formed = moles of CaCO₃ reacted = 0.066

mass of CaO formed = moles*molar mass = 0.066*56 = 3.696

% of solid carbon = (0.408/3.4+3.696)*100 = 5.76 %