Question

At 650 K, the reaction MgCO3(s)⇌MgO(s)+CO2(g) has Kp=0.026. A 10.0-L container at 650 K has 1.0 g of MgO(s) and CO2 at P=0.260 atm. The container is then compressed to a volume of 0.100 L. Find the Mass of MgCO₃ that forms.

Answer 1

MgCO3(s)   <---------------------> MgO(s)+CO2(g)

Kp = P_CO2

0.260 = P_CO2

moles of MgO = 1 / 40 = 0.025

from

P V = n R T

0.260 x 10 = n x 0.0821 x 650

n = 0.0487

moles of CO2 = 0.0487

MgCO3(s)   <---------------------> MgO(s)   +   CO2(g)

        1                                                 1                1

                                                     0.025          0.0487

here limiting reagent is MgO. so MgCO3 formed based on that.

1 mol of MgO -----------------> 1 mol of MgCO3

0.025 mol of MgO --------------->   ?? MgCO3

moles of MgCO3 = 0.025 x 1 / 1 = 0.025

mass of MgCO3 = moles x molar mass

                           = 0.025 x 84

                          = 2.1 g