Question

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?

Answer 1

For,

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

Now,

K_c = [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