Balanced equations: P2O5 + 3H2O → 2 H3PO4 ------- (I) H3PO4 (aq) + 2 NH4OH (aq) → (NH4)2HPO4 (s)+ 2H2O --------- (II) 1 mole of P2O5 (Phosphorus Pentoxide) with excess water gives 2 moles of Phosphoric acid H3PO4. 1 mole of Phosphoric acid in excess of Ammonium Hydroxide, produces 1 mole of (NH4)2HPO4 salt (s) and water.Molecular weight of P2O5 = 142Molecular weight of (NH4)2HPO4 = 132 Multiply the equation (II) by a factor of 2 we get 4NH4OH + 2H3PO4 → 2(NH4)2HPO4 + 4H2O …………….(III)Combining equations (I) & (III) P2O5 + 4NH4OH → 2(NH4)2HPO4 + H2O ……………………(IV)From equation (IV) 1 mole of P2O5 (142g) gives 2 moles of (NH4)2HPO4 (264 g) on hydration followed by neutralisation with excess of NH4OH.Mass of P2O5 required to produce 100 g of (NH4)2HPO4 = 142(100)/264 = 53.79 g
