When 0.62 g of BaCl2.2H20 reacts with excess Na3PO4.12H20, how many moles of Ba3(PO4)2 will be produced? (M.M for Na3PO4.12H20 = 380 g/mol, BaCl2.2H,0 = 244 g/mol, Baz(PO4)2 = 602 g/mole). 2 Na3PO4.12H20 + 3 BaCl2.2H20 - Bag(PO4)2 + 6 NaCl + 30 H20 a. 3.8*10* mole O b.7.6*10*mole O c. 8.5*10-4 mole O d. 1.7*10-3 mole