What is the maximum mass of Mg3(PO4)2 that can be produced?

Balance the equation

3 MgCl2(aq) + 2 Na3PO4(aq) ----> Mg3(PO4)2(aq) + 6 NaCl(aq)


Work out moles of each reagent

moles = molarity x litres
moles MgCl2 = 0.310 M x 0.0475 L = 0.014725 moles MgCl2

moles Na3PO4 = 0.480 M x 0.0915 L = 0.04392 moles

Balanced equation shows
3 moles MgCl2 needs 2 moles Na3PO4 to react
So 1 mole MgCl2 needs 2/3 moles Na3PO4
Thus  0.014725 moles MgCl2 needs (2/3 x 0.014725) moles Na3PO4
= 0.009817 moles Na3PO4

We need 0.009817 moles Na3PO4 to react with all the MgCl2 given, and we have 0.04392 moles Na3PO4, which is more then enough, so Na3PO4 is the excess reagent and MgCl2 is the limiting reagent.

The maximum amount of product possible is if all the limiting reagent reacts.
So work out moles of Mg3(PO4)2 produced from 0.014725 moles of MgCl2


3 moles MgCl2 reacts to form 1 mole Mg3(PO4)2
So 0.014725 moles MgCl2 will form 1/3 x 0.014725) moles Mg3(PO4)2
= 0.004908 moles Mg3(PO4)2 possible

mass = molar mass x moles
mass Mg3(PO4)2 = 262.87 g/mol x 0.004908 mol
= 1.29 g