Calcium carbonate, CaCO 3 , decomposes when heated to give calcium oxide, CaO, a

Question

Calcium carbonate, CaCO3, decomposes when heated to give calcium oxide, CaO, and carbon dioxide, CO2.

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

Kp for this reaction at 940°C is 1.216.

(a) What would be the yield of carbon dioxide (in grams) when 5.576 g of CaCO3 and 5.576 g CaO are heated to 940 °C in a 5.700 L vessel. (Ignore the volume occupied by the solids.)

Mass of CO2 = g

(b) What would be the effect of adding a similar quantity of carbon dioxide to this equilibrium mixture?

Mass of CO2 = g

Mass of CaO = g

Mass of CaCO3 = g

(c) What would happen if the quantity of calcium carbonate were doubled?

Mass of CO2 = g

Mass of CaO = g

Mass of CaCO3 = g

Explanation / Answer

CaCO3(s) = CaO(s) + CO2(g)

Kp = PCO2  

Kc = [CO2]

Kp = Kc*(R*T)n ; where n = moles of gaseous products - moles of gaseous reactants in the balanced reaction ; R = universal Gas Constant ; T = temperature in kelvin

Now., molar mass of CaCO3 = 100 g/mole

Thus, moles of CaCO3 in 5.576 g of it = mass/molar mass = 5.576/100 = 0.05576

Now, Applying Ideal Gas Equation; i.e. P*V = n*T*T; we get

moles of CO2 formed, n = (P*V)/(R*T)

Now, P = pressure of CO2 = Kp = 1.216 atm ; V = volume in litres = 5.7 L ; R =0.0821 ; T = 1213 K

or, n = (1.216*5.7)/(0.0821*1213) = 0.07

Thus, mass of CO2 formed = moles *molar mass = 0.07*44 = = 3.06 g ( molar mass of CO2 = 44 g/mole)

b) When CO2 is added, then backward reaction will occur and CaCO3 formation will take place until the pressure of CO2 reduces to 1.216 atm of mass of CO2 reduces back to the original

c) If the quantity of CaCO3 is doubled, there will not be any changes in the mass & pressure of CO2

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