Question

At 650 K, the reaction MgCO3(s)⇌MgO(s)+CO2(g) has Kp=0.026. A 14.4 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 0.800 L .Find the mass of MgCO3 that is formed.

Answer 1

The given reaction is,

MgCO₃(s) MgO(s) + CO₂(g)

Here, the equilbrium has only one gas i.e CO₂(g) which contributes to the K_p.

Now, K_p = [CO₂]

The moles of MgO can be found as---

(1.0 gram MgO) (1 mole MgO / 40.30 grams) = 0.0248 moles of MgO

Again , V = 14.4 L - 0.800 L = 13.6 L

We find moles that would need to be shifted to the left, into producing MgCO₃

PV = nRT

=> 0.0260 atm *13.6 L = n * (0.08206 L-atm/mol-K) (650K)

=> n = 0.3536 /53.339

=> n = 0.006629 moles of CO₂

And that will be the limiting reagent.

Since MgO(s) + CO₂(g) react in a 1:1 mole ratio, it is the CO₂'s moles that determine the amount of MgCO₃ that will be produced
By the equation:
MgCO₃(s) MgO(s) + CO₂(g)
0.006629 moles of CO₂ produces an equal 0.006629 moles of MgCO₃

Finally, using molar mass,
(0.006629 moles of MgCO₃) (84.3139 g MgCO3 /1 mole ) = 0.558 grams of MgCO₃