Asked by Kevin
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?
Answers
Answered by
DrBob222
Convert 3.9565 g Ca3(PO4)2 to moles, then to moles CaCl2, then M = moles CaCl2/L CaCl2.
Answered by
Kevin
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
Answered by
DrBob222
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.