Asked by Rory
When heated, CaCO3 decomposes as follows: CaCO3 (s) = CO2 (g) +CaO (s)
in an experiment, 10.00 g of CaCO3 was placed in a .500 L container, and when it was heated to a constant temperature, 50% of the solid decomposed. What is the value of Kc for this reaction under these conditions?
in an experiment, 10.00 g of CaCO3 was placed in a .500 L container, and when it was heated to a constant temperature, 50% of the solid decomposed. What is the value of Kc for this reaction under these conditions?
Answers
Answered by
DrBob222
Kc = (CO2)
If we start with 10.00 g CaCO3 and half of it decomposes, that is 5.00 grams.
5 g CaCO3 = g/molar mass = 5.00/100 = 0.05 moles CaCO3. That will give, from the equation, 0.05 moles CO2 and the concn is moles/L; therefore, (CO2) = 0.05/0.5 L = 0.1 M. Thus Kc = 0.1.
If we start with 10.00 g CaCO3 and half of it decomposes, that is 5.00 grams.
5 g CaCO3 = g/molar mass = 5.00/100 = 0.05 moles CaCO3. That will give, from the equation, 0.05 moles CO2 and the concn is moles/L; therefore, (CO2) = 0.05/0.5 L = 0.1 M. Thus Kc = 0.1.
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