1. 3 Fe(OH)2 + 2 H3PO4 → Fe3(PO4)2 + 6 H2O
double displacement
Fe2+
PO43-
PO43- H+
Fe2+
OH-
Fe2+ H+
PO43- H+
OH-
OH-
Fe2+ H+
3 Fe(OH)2 + 2 H3PO4 → Fe3(PO4)2 + 6 H2O
iron (II) hydroxide
phosphoric acid
iron (II) phosphate
water
1 mol Fe(OH)2
2 mol H3PO4
97.99 g H3PO4
3.2 g Fe(OH)2 x x x =
g Fe(OH)2
3 mol Fe(OH)2
1 mol H3PO4
2.33 g H3PO4
H3PO4 is in excess
Given: 2.50 g H3PO4
Thus, Fe(OH)2 is the limiting reagent =
0.17 g H3PO4 in excess

2. 3 Fe(OH)2 + 2 H3PO4 → Fe3(PO4)2 + 6 H2O
1 mol Fe(OH)2
1 mol Fe3(PO4)2
g Fe3(PO4)2
3.2 g Fe(OH)2 x x x =
g Fe(OH)2
3 mol Fe(OH)2
1 mol Fe3(PO4)2
4.23 g Fe3(PO4)2 formed
1 mol Fe(OH)2
6 mol H2O
g H2O
3.2 g Fe(OH)2 x x x =
g Fe(OH)2
3 mol Fe(OH)2
1 mol H2O
1.28 g H2O formed
% Yield =
Actual Yield
Theoretical Yield
x 100
3.99
% Yield =
x 100
94.3% =
4.23