Reaction:
P2O5 + 3H2O → 2 H3PO4 - (I)
H3PO4 (aq) + 2 NH4OH (aq) → (NH4)2HPO4 (s)+ 2H2O - (II)
Atomic weight of P2O5 = 142
Atomic weight of (NH4)2HPO4 = 132
Increase the condition (II) by a variable of 2 we get
4NH4OH + 2H3PO4 → 2(NH4)2HPO4 + 4H2O … .(III)
from eqns. (I) and (III)
P2O5 + 4NH4OH → 2(NH4)2HPO4 + H2O … (IV)
From condition (IV) 1 mole of P2O5 (142g) gives 2 moles of (NH4)2HPO4 (264 g) on hydration taken after by balance with overabundance of NH4OH.
Mass of P2O5 required to create 100 g of (NH4)2HPO4 = 142(100)/264 = 53.79 g
