posted by Jody on .
If 0.0142 M aqueous Cl- is produced stoichiometrically according to the balanced equation in a reaction solution with a total volume of 1480 mL, how many mol of solid Ca3(PO4)2 are produced?
3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCl(aq)
How many moles Cl^- are produced? That will be 0.0142M x 1.480L = ?
Convert that to moles Ca3(PO4)2.
?mol NaCl x (1 mol Ca3(PO4)2/6 moles NaCl) = ?mols NaCl x (1/6) = x mols Ca3(PO4)2.