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CaCO3(s) <------> CaO(s) + CO2(g) Kc = 0.10 at some temperature How many grams of CO2...

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?

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Answer #1

For,

CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)                    Kc = 0.10

Now,

Kc = [CO2]                 (\because 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

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