Consider the reaction:
\(\mathrm{CaCO}_{3} \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)
A sample of \(\mathrm{CaCO}_{3}\) is placed into a sealed 0.500-L container and heated to 550 K at which the equilibrium constant is \(4.1^{*} 10^{-4}\). When the reaction has come to equilibrium, what mass of solid CaO is in the container? (Assume that the sample of \(\mathrm{CaCO}_{3}\) was large enough
that equilibrium could be achieved.)
Equation Transcription:
CaCO3
4.1 * 10-4
Text Transcription:
CaCO_3 Harpoon CaO(s) + CO_2(g)
CaCO_3
4.1 times 10^-4
Solution:
Step 1
The equilibrium constant is
Keq =
It is known that the equilibrium constants for pure solids and liquids are zero. So that [CaCO3(s)] and [CaO(s)] will be equal to zero. The above equation can be written as,
Keq = [CO2]
Here it has been given that Keq = 4.1 × 10−4
Thus Keq = [CO2] = 4.1× 10−4 = 0.00041M
The mole of CaO = 0.5 L× 0.00041 mol/L= 0.000205 mol
Molar mass of CaO = 56.07g/mol
Thus mass of CaO in the container = 56.07 g/mol × 0.000205 mol = 0.0115g
Thus the container have 0.0115 g of CaO.
