Question

52.8 mL of a calcium chloride solution with unknown concentration was treated with phosphoric acid to remove all of the calcium ions in the form of a precipitate. The precipitate was filtered, dried and was found to have the mass of 3.9565 g. What was the concentration of the calcium chloride solution in M?

Answer 1

Convert 3.9565 g Ca3(PO4)2 to moles, then to moles CaCl2, then M = moles CaCl2/L CaCl2.

Answer 2

i got the moles of Ca3(PO4)2=0.01275549 but how do i covert that to moles of CaCl2? using the stoichiometric ratio which is 1:1 or am i doing something wrong

Answer 3

Yes and no. Yes, you use the stoichiometric ratio but it isn't 1:1 is it? I see 3 moles CaCl2 = 1 mole Ca3(PO4)2.