CaCO3 (s) ⇄ CaO (s) + CO2 (g) 0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.270 atm of CO2 is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO3? 15.01 g

11.58 g

Explanation:

The equilibrium constant based on pressure (Kp) only relates the gases, so, for the reaction given, only the pressure of CO₂ will be considered. The equation is already balanced, so

Kp = pCO₂

Kp = 0.220

So, if 0.270 atm is added, the equilibrium will shift to form the reactant, and Kp will remain the same (0.220), so the gas added will form CaCO₃. The number of moles of CO₂ can be calculated by the equation of ideal gas:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the constant of the gases (R = 0.082 atm. L/mol. K), and is the temperature.

First, let's calculated how much moles of CO₂ were formed in the equilibrium:

0.220x10 = nx0.082x385

31.57n = 2.20

n = 0.0697 mol

The stoichiometry is 1 mol of CaCO₃ for 1 mol of CO₂, so it was consumed 0.0697 mol of CaCO₃, and in the equilibrium, the number of moles of it is 0.100 - 0.0697 = 0.0303 mol

For the quantity added, the number of moles is:

0.270x10 = nx0.082x385

31.57n = 2.70

n = 0.0855 mol

This will form more 0.0855 mol of CaCO₃, so the final number of moles is:

n = 0.0303 + 0.0855 = 0.1158 mol

The molar masses are: Ca = 40 g/mol, C = 12 g/mol, O = 16 g/mol

CaCO₃ = 40 + 12 + 3x16 = 100 g/mol

The mass is the number of moles multiplied by the molar mass:

m = 0.1158x100

m = 11.58 g