# P2O5 on treatment with excess H2O followed by excess of NH4OH

P2O5 on treatment with excess H2O followed by excess of NH4OH form (NH4)2HPO4. If 100 g of (NH4)2HPO4 is formed then find out the mass of P2O5 initially taken

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 = 142
Molecular weight of (NH4)2HPO4 = 132

Multiply the equation (II) by a factor of 2 we get
4NH2OH + 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