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
