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Gas mixture problem!

  1. Oct 18, 2006 #1
    I spent quite a bit of time trying to figure this out, but I just can't seem to get this question:

    When a 10.0g mixture of Ca(ClO3)2 abd Ca(ClO)2 is heated to 973 Kelvins in a 10.0 L vessel, both compounds decompose, forming O2 and CaCl2. The final pressure inside the vessel is 1.00 atm. What is the mass of each compount in the original mixture?

    What I have done:
    Used ideal gas law to calculate the number of moles in final mixture of gas. Got ~0.125 moles

    I then let (x) be the initial grams of Ca(ClO3)2 and (x-10) be the initial grams of Ca(ClO)2. I wrote the balanced eq'ns for the decomposition for both.

    Ca(ClO3)2 --> 3(O2) + CaCl2
    Ca(ClO)2 --> O2 + CaCl2

    • If (x)/(molar mass of Ca(ClO3)2)= number of moles of Ca(ClO3)2
      Let 3(x)/(molar mass of Ca(ClO3)2)= number of moles of O2 in first decomposition
      Let (x)/(molar mass of Ca(ClO3)2)= number of moles of CaCl2
    For the 2nd eq'n
    If (10-x)/(molar mass of Ca(ClO)2) = number of moles Ca(ClO)2
    Let (10-x)/(molar mass of Ca(ClO)2) = number of moles O2 in 2nd decomposition
    Let (10-x)/(molar mass of Ca(ClO)2) = number of moles CaCl2 in 2nd decomposition.

    (10-x)/(molar mass of Ca(ClO)2) + (10-x)/(molar mass of Ca(ClO)2) + 3(x)/(molar mass of Ca(ClO3)2) + (x)/(molar mass of Ca(ClO3)2) = 0.125 moles.

    I was guessing if I solved for x, I would be able to get the amount of grams...but I keep getting a negative value. I'm not sure why this doesn't work, it seems to make enough sense to me.
  2. jcsd
  3. Oct 18, 2006 #2
    bump b/c this question owns me
  4. Oct 19, 2006 #3


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    Staff: Mentor

    Check your equation - solids are not gaseous ;)
  5. Oct 20, 2006 #4


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