Question

At 650 K, the reaction MgCO3(s)<->MgO(s)+CO2(g) has Kp=0.026. A 10.1L container at 650 K has 1.0g of MgO(s) and CO2 at P = 0.0260 atm. The container is then compressed to a volume of0.500L . Find the mass of MgCO3 that is formed.

Answer 1

the equilibriun has only one gas that contributes to the Kp, that's the CO2:
MgCO3(s) <=> MgO(s) + CO2(g)

Kp = [CO2]

===============

find moles, using molar mass:
1.0 grams Mg) (1 mole MgO / 40.30 grams) = 0.0248 moles of MgO

LeChatelier's principle would entertain the possibliity of shifting the moles of CO2 that had occupied the lost 10.0 Litres & still be able to maintain the 0.026atm & 650 K, which is the only way it could maintain 0.0260 atm of CO2, & its Kp of 0.0260.....
find moles that would need to be shifted to the left, into producing MgCO3......
PV = nRT
(0.0260 atm) (10.0 L) = n (0.08206 L-atm/mol-K) (650K)
n = 0.00487 moles of CO2
that's the limiting reagent

ince MgO(s) + CO2(g react in a 1mole to 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)
0.00487 moles of CO2 produces an equal 0.00487 moles of MgCO3


using molar mass:
(0.00487 moles of MgCO3) (84.31 g MgCO3 / mole ) = 0.41096 grams of MgCO3