Answer:
The answer to your question is: kc = 6.48
Explanation:
Data
Given Molecular weight
CaO = 44.6 g 56 g
CO₂ = 26 g 44 g
CaCO₃ = 42.3 g 100 g
Find moles
CaO 56 g ---------------- 1 mol
44.6 g -------------- x
x = (44.6 x 1) / 56 = 0.8 mol
CO₂ 44 g ----------------- 1 mol
26 g ---------------- x
x = (26 x 1 ) / 44 = 0.6 moles
CaCO₃ 100 g --------------- 1 mol
42.3g -------------- x
x = (42.3 x 1) / 100 = 0.423 moles
Concentrations
CaO = 0.8 / 6.5 = 0.12 M
CO₂ = 0.6 / 6.5 = 0.09 M
CaCO₃ = 0.423 / 6.5 = 0.07 M
Equilibrium constant = [tex]\frac{[products]}{[reactants]}[/tex]
Kc = [0.07] / [[0.12][0.09]
Kc = 0.07 / 0.0108
kc = 6.48
