posted by gina .
For the reaction below, Kp = 1.16 at 800.°C.
CaCO3(s) CaO(s) + CO2(g)
If a 21.7-g sample of CaCO3 is put into a 10.6 L container and heated to 800°C, what percentage by mass of the CaCO3 will react to reach equilibrium?
Kp = 1.16 = pCO2
How many mols CO2 can be formed in a 10.6L container @ 800 C.
n = PV/RT = 1.16*10.6/0.08206*1073 = about 0.14 but you need to do it more accurately. You had 21.7/100 = 0.217 mols.
% decomposed = (0.14/0.217)*100 = ?