3MgCl2(aq) + 2Na3PO4(aq) ------> 6NaCl(aq) + Mg3(PO4)2(s)
no of moles of MgCl2 = W/G.M.Wt
                                   Â
=29/95Â Â = 0.305 moles
no of moles of Na3PO4 = W/G.M.Wt
                                      Â
= 29/164 = 0.177 moles
3 moles of MgCl2 react with 2 moles of Na3PO4
0.305 moles of MgCl2 react with 2*0.305/3Â Â Â =
0.203 moles ofNa3PO4
Na3PO4 is limiting reagent
2 moles of Na2PO4 react with MgCl2 to gives 1 moles of
Mg3(PO4)2
0.177 moles of Na2PO4 react with MgCl2 to gives = 1*0.177/2 =
0.0885 moles of Mg3(PO4)2
mass of Mg3PO4)2 = no of moles * gram molar mass
                               Â
= 0.0885*262.87Â Â =23.26g
