Given the reaction 3 H3PO4 + 2 Ba(OH)2 ⇒ BaPO4 + 6 H20
How many grams of precipitate could you make if you reacted 286.0 mL of 3.000 M H3PO4 with 855.0 mL of 1.400 M Ba(OH)2

240.17 g Ba3(PO4)2

Explanation:

1. Determine the limiting reactant.

2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O

moles H3PO4 = M x V = 3 x 0.286 = .858 moles H3PO4

moles Ba(OH)2 = M x V = 1.4 x 0.855 = 1.197 moles Ba(OH)2

ratio Ba(OH)2 : H3PO4 = 1.197: .858 = 1.39: 1

stoichiometric ratio Ba(OH)2 : H3PO4  = 3:2

Ba(OH)2is the limiting reactant

MM Ba3(PO4)2 = 601.92 g/mol

g Ba3(PO4)2 = moles Ba(OH)2 x(1 mol Ba3(PO4)2/3 moles Ba(OH)2) x (MM Ba3(PO4)2/ 1mol Ba3(PO4)2) = 1.197 x 1/3 x 601.92 = 240.17 g Ba3(PO4)2