What is the mass of each compound in the original mixture when heated to 700C?

[leolaw]

I am stuck in this gas problem:
You are given a 10.00 gram solid mixture of $$Ca(ClO_3)_2$$ and $$Ca(ClO)_2$$. When this mixture is heated, both componds decomplse, releasing oxygen gas and leaving behind solid $$CaCl_2$$

When this 10.00gram mixture is sealed in a 10.0-Liter evacuated vessel and heated to 700C, both compounds completely decompose according to the above reactions. The final pressure in the vessel is 1.00atm. Determine the mass of each compound in the original mixture.

Ok. I have written out the balanced equation of the two substances:
$$Ca(ClO_3)_2 + O_2 --> CaCl_2 + 4O_2$$ and
$$Ca(ClO)_2 + O_2 --> CaCl_2 + 2O_2$$

using the ideal gas law, i have found that the total $$O_2$$ collected is 0.1252 mol (V = 10L , T = 973K, P = 1.00atm)

so I have let x equal to the gram of $$Ca(ClO_3)_2$$ and y equal to the gram of $$Ca(ClO)_2$$, combine them toegether, I have:
x + y = 10

and then i found out the precentage of O2 that each substance produce, so I have another equation:
(0.0193) x + (0.0186) y = 0.1252

when i solve for x and y , i have x equal to -86 and y = 96, which obviously, is not possible?

Any clues on how to solve this probelm?

 

[gravenewworld]

From the way the problem is worded I think you equations should be

$$Ca(ClO_3)_2 --> CaCl_2 + 3O_2$$
$$Ca(ClO)_2 --> CaCl_2 + O_2$$

Lets assume you have 1 mol of each starting material. Then according to the balanced equations, after decomposition Ca(ClO3)2 will give off 3 moles of O2 and Ca(ClO)2 will give off 1 mole of O2 for a total of 4 moles of oxygen. In other words 3/4 of the total number of moles come from Ca(ClO3)2 and 1/4 of the total number of moles of O2 come from Ca(ClO)2. You found .1252mol of total oxygen, so 3/4th of that would be .0939 moles of O2 coming from Ca(ClO3)2 and 1/4ths of that would be .0313 moles of O2 that would be coming from Ca(ClO)2.


Using the balanced equations again-

0.0939 mol O2 x 1molCa(ClO3)2/3mol O2=.0313 mol Ca(ClO3)2
and
.0313 mol O2 x 1molCa(ClO)2/1mol O2=0.0313 mol Ca(ClO)2

Now to double check that you are right you can add up the moles of each substance in grams and you should get 10.0g.

mw of Ca(ClO3)2 206.98g/mol
and mw of Ca(ClO)2 is 142.98 g/mol

thus .0313 mol(206.98g/mol)+.0313mol(142.98g/mol)=10.0g (approximately due to rounding). So you know you have to be right

 

[leolaw]

Yes, I have fixed up on the equation.
THx