Calculate the moles of MgCl2 that will be required to produce 535 grams of Mg3(PO4)2 from MgCl2(aq) + Na3PO4(aq) -> Mg3(PO4)2(s) + NaCl(aq).
a) 4.17 mol MgCl2
b) 2.78 mol MgCl2
c) 5.56 mol MgCl2
d) 3.94 mol MgCl2