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10 g of a mixture of CaCl2 and NaCl is treated to precipitate all the calcium as CaCO3. This CaCO3 is heated to convert all the Ca to CaO and the final mass of CaO is 1.68 g. The percentage by mass of CaCl2 in the original mixture is what?

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The reactions are: 
 CaCl2 + Na2CO3  = CaCO3  + 2NaCl       
 CaC03 + heat = CaO + CO2 

1 mole of CaO is formed from 1 mole CaCO3 
1 mole of CaCO3 is formed from 1 mole of CaCl2 
Mole ratio of CaO to CaCl2 is 1: 1 
Moles of CaO formed = mass/molar mass = 1.68/56=0.03mol 
Moles of CaCl2 = 0.03mol 
Mass of CaCl2 = 0.03 * 111 = 3.33g 
The total mass of the mixture of calcium chloride(CaCl2) and sodium chloride(NaCl) given as 10g,
So, the % mass of CaCl2 in the original mixture is  = 33.3 %

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