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Thermal decomposition of calcium carbonate - problem

  1. Feb 11, 2014 #1
    Hello, here goes the problem:
    The mass of calcium carbonate upon thermal decomposition decreased by 1/5. (a) How many molecules of CaCO3 per 100 molecules were decomposed to CaO and CO2. (b) The content of CaO in the final sample express in molar fraction.

    My solution:
    I set the starting sample to 100g which corresponds to 1 mole of CaCO3. The decrease of mass by 1/5 corresponds to 20g or 0.2 mole. So the solution seems to me to be 20 molecules per 100 molecules.
    Ad (b) If 0.2 mole of CaCO3 were decomposed, 0.2 mole of CaO and 0.2 mole CO2 should be formed. Thus the molar fraction of CaO in final samples should be 0.2/(0.2+0.2+0.8) = 1/6.

    Yet the provided solutions from two scources differs. Where, if so, did I make mistake?
    Thank you for your help.
     
  2. jcsd
  3. Feb 11, 2014 #2

    Borek

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    Staff: Mentor

    OK

    OK

    No. Mass decreased by 20 g, but it doesn't mean 20 g decomposed and disappeared.

    Write reaction equation - what is lost during decomposition? What is left?

    And please, don't ignore the template.
     
  4. Feb 11, 2014 #3
    CaCO3 → CaO + CO2

    Doesn't 20 g of the decomposed reactant corresponds to decomposition of 0.2 mole of that reactant? I don't assume that mass dissappeares. It appears as 20 g of products (0.4 moles of them). So now we are left with 0.8 moles of CaCO3, isn't that true?

    Ok.
     
    Last edited: Feb 11, 2014
  5. Feb 11, 2014 #4

    Borek

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    Staff: Mentor

    I already asked you - what is lost, what is left? Add state symbols to all three substances and it should become obvious.
     
  6. Feb 11, 2014 #5
    Should I assume that 20 g went away as CO2? So from 1 to 1 mole ratio, 0.45 moles of CaCO3 decomposed, which corresponds to 45 molecules per 100 molecules.
    We are left with 0.45 moles of CaO and 0.55 moles of CaCO3 which is 45 % of CaO in the final sample.
    Ok, the solutions agree.
    Thank you, Borek, for making me think and approaching the problem from more real and "chemical point of view".
     
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