Find the mass in the original mixture:

Question

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?

Answer 1

The reactions are: 
 CaCl₂ + Na₂CO₃  = CaCO₃  + 2NaCl       
 CaC0_{3 +} 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 %