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 

Asked by Anil | 11th May, 2017, 12:04: PM

Expert Answer:

Balanced equations:
    P2O5 + 3H2O  → 2 H3PO -------  (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  
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 


Answered by Prachi Sawant | 11th May, 2017, 05:00: PM