[Swill175]

Both calcium chloride, CaCl₂, and sodium chloride, NaCl, are used to melt ice and snow on roads in the winter. I hired a company to prepare a mixture of these two compounds for this purpose. As a chemist, I decided to check their formulation. I did this by dissolving 2.463 g of the mixture in water and precipitated the calcium by adding sodium oxalate, Na₂C₂O₄

Ca²⁺ (aq) + C₂O₄^2- (aq) --> CaC₂O₄ (s)

The calcium oxalate was then carefully filtered from the solution, dissolved in sulfuric acid (this converts the CaC₂O₄ to H₂C₂O₄) and titrated with 0.1000 M KMnO₄. The reaction that occurs here is

6H⁺ + 5H₂C₂O₄ + 2MnO₄^- --> 10CO₂ + 2Mn²⁺ + 8H₂O

The titration required 21.62 mL of the KMnO₄ solution.


1. How many moles of C₂O₄^2- are present in the CaC₂O₄ precipitate?

2. How many grams of CaCl₂ are in the original 2.463 g sample used?

3. What is the percentage by mass of CaCl₂ in the sample used.

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  So confused Thanks!!

[agrobert]

I don't understand what is confusing.  These are stoichiometric relationships.  You should try the problem before expecting people to do your calculations.  Did you standardize your KMnO₄?

[AWK]

Start from calculation of moles KMnO4

[Borek]

Did you standardize your KMnO₄?

0.1000 M, given in the question. Don't add to Swill175 confusion.

[Swill175]

Start from calculation of moles KMnO4

I still don't understand - if I find the amount of moles titrated w/ KMNO₄..how would i relate that back to C₂O₄^2- or even CaCl₂? 

[AWK]

Start from calculation of moles KMnO4

I still don't understand - if I find the amount of moles titrated w/ KMNO₄..how would i relate that back to C₂O₄^2- or even CaCl₂? 

Then use  stoichiometry of reaction:
6H+ + 5H2C2O4 + 2MnO4- --> 10CO2 + 2Mn2+ + 8H2O
(which is corretly balanced)

moles H2C2O4 are equal to moles CaC2O4 which are on turn equal to moles of CaCl2