Question

At 650 K, the reaction MgCO3(s)⇌MgO(s)+CO2(g) has Kp=0.026.

A 14.5 L container at 650 K has 1.0 g of MgO(s) and CO2 at P = 0.0260 atm. The container is then compressed to a volume of 0.900 L .

Find the mass of MgCO3 that is formed.

Answer 1

For MgCO₃(s)⇌MgO(s)+CO₂(g) , Kp = P_co2, since CO₂ is the only gaseous compound in this reaction.

Therefore, P_co2 = 0.0260atm

Now , Volume , V = 14.5L , Temperature,T = 650K , Pressure , P = 0.0260atm

Using Ideal gas Equation: PV = nRT , R = 0.08206 L atm mol^-1 K^-1

so. no. of moles of CO₂, n = PV/RT = (0.0260atm)(14.5L)/(0.08206 L atm mol^-1 K^-1)(650K) no. of moles of CO₂, n  = 0.007 mol

Molar mass of MgCO₃ = 40.3 g/mol , mass of MgCO₃ = 1g

no. of moles of MgCO3 = 1g / (40.3g/mol) = 0.025 mol

Now, considering the reverse reaction(i.e formation of MgCO₃) : MgO + CO₂  ------> MgCO₃

so, 1 mole of MgO and 1 mole of CO2 reacts to give 1 mole of MgCO₃ . We have 0.025 mol of MgO and 0.007 mol of CO₂ . Since amount of CO₂ is limited , therefore it is limiting reagent and amount of product i.e MgCO₃ depends on CO₂. Hence, 0.007mol of CO₂ will react with 0.007 MgO to give 0.007 mole of MgCO₃.

Mass of MgCO₃ formed = 0.007mole X molar mass of MgCO₃ = 0.007mole X 84.3g/mol = 0.5901g

Mass of MgCO₃ formed = = 0.5901g