95.0g CaO = 1.70 moles
[Kc = 0.0108 = 0.0108 M CO2 in 50 L = 0.54 moles CO2]
CaCO3 => CaO + CO2
I 6.55 1.70 1.32
C +x -x -x
E 6.55+x 1.70-x 1.32-x
Plugging ONLY the CO2 data into the Kc expression (because solids cannot be included), we now get:
1.32 moles -x = 0.54
x = 0.78 moles changed
and this amount is used to change the CO2, CaO and CaCO3 numbers…..
3a) Final answers for part (a) CaCO3 = 6.55 moles + 0.78 moles = 7.33 moles = 733g
CaO = 1.70 moles – 0.78 moles = 0.92 moles = 51.5 g
CO2 = 0.0264 – 0.0156 = 0.0108 M…Pco2 = 1.04 atm or 23.76 grams
1a) The same logic is used to begin part (b) as part (a)…this time the reaction is shifting RIGHT because the given PCO2 of .104 is less than Kp.
1b) Fill in the ICE chart:
780g CaCO3 = 7.80 moles
31.0g CaO = .554 moles
P(CO2) = 0.104 or Pco2 = .104 atm = .0540 moles
CaCO3 => CaO + CO2
I 7.80 .554 0.0540
C -x +x +x
E 7.8 - x .554+x 0.0540+x
Solving for x….
0.0540+x = 0.54
x = 0.486 moles changed
3b) Final answers for part (b) CaCO3 = 7.80 moles - 0.486 moles = 2.90 moles = 290g
CaO = .554 + 0.486 = 1.04 moles = 58.2g
CO2 = 0.0540 + 0.54 moles = 23.76 g