CaCO3(s) <------> CaO(s) + CO2(g) Kc = 0.10 at some temperature How many grams of CO2 are formed when equilibrium is established at that temperature in a 1.0-L container?
For,
CaCO3(s) CaO(s) + CO2(g) Kc = 0.10
Now,
Kc = [CO2] ( concentration of solids do not not appear in the equilibrium constant expression)
or, [CO2] = 0.10 M
Volume of the container = 1.0 L
Moles of CO2 formed at equilibrium = 0.10 M x 1.0 L = 0.10 mol
Molar mass of CO2 = 44 g/mol
Hence, the mass of CO2 formed at equilibrium = 0.10 mol x 44 g/mol = 4.4 g
CaCO3(s) <------> CaO(s) + CO2(g) Kc = 0.10 at some temperature How many grams of CO2...
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