Hard water often contains dissolved ca2+ and mg2+ ions. one way to soften water is to add phosphates. the phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. suppose that a solution is 5.5×10−2m in calcium chloride and 9.5×10−2m in magnesium nitrate.

what mass of sodium phosphate must be added to 2.0l of this solution to completely eliminate the hard water ions? assume complete reaction.

Ok so, Im going to try to explain this: What we need to do we are going to do this on 3 steps: 1) We need to get the total of moles of calcium + magnesium. 2) Multiply it by 2/3 to get the moles of phosphate 3) from there we need to get the  grams of sodium phosphate. So we need those three operations like this:

Moles calcium = 5.5 x 10^-2 moles/liter x 2 liters = 11.0 x 10^-2 moles 
Moles magnesium = 9.5 x 10^-2 moles/liter x 2 liters = 19.0 x 10^-2 moles 
Total moles of Ca + Mg = 11.0 x 10^-2 + 19.0 x 10^-2 = 30.0 x 10^-2 moles. 

Then we include the grams of Na3PO4 that is  163.9408 g/mol and finally you do this operation 163.9408 g/mol x 20.0 x 10^-2 moles = 32.8 grams = 33 grams. And that is how you get it

Ibelieve your answer is true