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?

3 answers

Convert 3.9565 g Ca3(PO4)2 to moles, then to moles CaCl2, then M = moles CaCl2/L CaCl2.
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
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.
Similar Questions
  1. 3ca(s) + 2H3PO4(aq) -> Ca3(PO4)2(aq)+3H2(g)1)a reaction between calcium and a solution of 1.2moldm-3 phosphoric acid produced 3
    1. answers icon 1 answer
    1. answers icon 1 answer
    1. answers icon 5 answers
  2. 2NaF + CaCl2 �¨ CaF2 + 2NaClAn excess of 0.20 M NaF is used to precipitate the calcium ions from a 50.0 mL sample of a calcium
    1. answers icon 4 answers
more similar questions