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. 0.290 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?

12.22 g

Explanation:

The reaction is

CaCO₃(s)⇄ CaO (s) + CO₂ (g)

The constant of equilibrium based on pressure (Kp) depends only on the gas substances, so in this case, it will depend on CO₂.

Kp = pCO₂ = 0.220

For the Le Chatêlier principle, a change in the pressure will shift the equilibrium to equilibrate the effect. Kp only changes with the temperature, so all the CO₂ added will form CaCO₃.

The stoichiometry is 1 mol of CaCO₃ for 1 mol of CO₂, and the number of moles of CO₂ can be calculated by the ideal gas equation:

PV = nRT

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

So, the number of moles of CO₂ in equilibrium is:

0.220x10 = nx0.082x385

31.57n = 2.20

n = 0.0697 mol

Which is the number of moles of CaCO₃ that reacts, so in the equilibrium, there is 0.1 - 0.0697 = 0.0303 mol of CaCO₃.

The number of moles of CO₂ added is:

0.290x10 = nx0.082=385

31.57n = 2.90

n = 0.09186 mol

This will totally form 0.09186 mol of CaCO₃, and it will have, in the final:

n = 0.0303 + 0.09186 = 0.1222 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 moles multiplied by the molar mass:

m = 0.1222 x 100

m = 12.22 g

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