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 0.200 L .

Find the mass of MgCO3 that is formed.

Answer 1

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

PV = nRT
(0.0260 atm) (14.5 L) = n (0.08206 L-atm/mol-K) (650K)
n = 0.007067 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)
0.007067 moles of CO2 produces an equal 0.007067 moles of MgCO3

using molar mass:
(0.007067 moles of MgCO3) (84.31 g MgCO3 / mole ) = 0.5959 grams of MgCO3