Question

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. 0.250 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?

Answer 1

Answer:

The final mass of CaCO3 is 10.68 grams

Explanation:

Step 1: Data given

Number of moles CaCO3 = 0.100 moles

Number of moles CaO = 0.100 moles

Volume = 10.0 L

When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.250 atm of CO2 is added, while keeping the temperature constant

Step 2: The balanced equation

CaCO3(s) <==> CaO(s) + CO2(g)

Step 3: Calculate moles of CO2

n = PV/RT

⇒n = the initial number of moles CO2 = TO BE DETERMINED

⇒P = the pressure of CO2 at theequilibrium = 0.220 atm

⇒V = the volume of the container = 7.0 L

⇒R = the gas constant = 0.08206 L*atm / mol * K

⇒T = the temperature = 385 K

n = 0.220*7.0/(0.08206*385) = 0.0487 (mol)

this is the amount of CaCO3 which has been converted to CaO before pumping-in additional 0.225 atm CO2(g).

Step 4: Calculate moles CaCO3

After adding additional 0.250 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm.  All this additional CO2 would completely convert to CaCO3:

n = PV/RT = 0.250*7.0/(0.08206*385) = 0.0554 moles

The total CaCO3 after equilibrium is reestablished is:

0.100 - 0.0487+ 0.0554 = 0.1067 mol

Step 5: Calculate mass CaCO3

Mass CaCO3 = 0.1067 moles * 100.09 g/mol

Mass CaCO3 = 10.68 grams

The final mass of CaCO3 is 10.68 grams