Question

( 2left(mathrm{NH}_{4}right)_{2} mathrm{HPO}_{4} Rightarrow mathrm{P}_{2} mathrm{O}_{5} )
2 moles of diammonium hydrogen phosphate, yield 1 mole of ( mathrm{P}_{2} mathrm{O}_{5} )
The molar mass of ( left(mathrm{NH}_{4}right)_{2} mathrm{HPO}_{4} ) is given by ( = )
( =36+1+31+44 )
( =132 mathrm{g} )
Molar mass of ( mathrm{P}_{2} mathrm{O}_{5} ) is given by ( = )
( =62+80 )
( =142 mathrm{g} )
So, percentage of ( mathrm{P}_{2} mathrm{O}_{5} ) is
( =[142 /(2) times 132] times 100 )
( =142 / 264 times 100 )
( =53.78 % )
Therefore percentage of ( mathrm{P}_{2} mathrm{O}_{3} ) is ( 53.78 % )

# Doust (a) 23.48 eace occuit of t (d) 1 da) 4 (6) 3 (c) 2. The percentage of P,0, in diammonium hydrogen phosphate, ((NH4)2 HPO4] is: (b) 46.96 (c) 53.78 (d) 71.00 of nitrogen in urea NH.CONH) is:

Solution