2K3PO4(aq) + 3MgBr2(aq) ===> Mg3(PO4)2(s) + 6KBr(aq) ... balanced equation
Finding limiting reactant:
moles K3PO4 present = 97.17 ml x 1 L/1000 ml x 1.4 mol/L = 0.1360 moles
moles MgBr2 present = 45.25 ml x 1 L/1000 ml x 0.615 mol/L = 0.02783 moles
Easy way to find limiting reactant is to divide moles by coefficient in balanced equation and see which is less:
For K3PO4 we have 0.1360 / 2 = 0.068
For MgBr2 we have 0.02783 / 3 = 0.0093 <== LIMITING
So, now we know that MgBr2 is limiting, meaning that K3PO4 is in excess. The question asks for the concentration of K3PO4 after the reaction.
moles K3PO4 used up during reaction: 0.02783 mol MgBr2 x 2 mol K3PO4/3 mol MgBr2 = 0.018555 moles
of K3PO4 used up.
Moles K3PO4 left over = 0.1360 moles - 0.01856 moles = 0.1174 moles K3PO4 left over.
What is the final volume? 97.17 ml + 45.24 ml = 142.41 mls = 0.14241 L
Final [K3PO4] = 0.1174 moles K3PO4 / 0.1424 L = 0.82 M (to 2 significant figures)