Solving Gas Mixture Problem: 10g Ca(ClO3)2 + Ca(ClO)2

[leaf345]

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.

Then,
(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.

 

[leaf345]

bump b/c this question owns me

 

[Borek]

Check your equation - solids are not gaseous ;)

 

[GCT]

Here's something from PF archives

https://www.physicsforums.com/archive/index.php/t-81537.html