[ 33 ] 



2 (NH 4 ) 3 P0 4 +2MgSO 4 =2MgNH 4 P0 4 + 2 (NH 4 ) 2 SO 4 

 2 MgNH 4 PO 4 (on ignition) ==Mg 2 P 2 O 7 +2NH 3 + H 2 O 

 Mg 2 P 2 O 7 : CaH 4 P 2 O 8 ::i9% : percentage of CaH 4 P 2 O 8 



iri the soluble portion. 

 i.e., 222 : 234:: 19 : percentage of Monocalcium phosphate 



present. 

 /. The percentage of Monocalcium phosphate in i gram 



of super = 



(8) To calculate the Tricalcium phosphate 



The total Mg 2 P 2 Oz came partly frohi the Mono and 



partly from the Tricalcium phosphate. 

 .*. Taking 19 from 21-5, 2*5% is left as the amount from 



the Tricalcium phosphate. 



Now, Ca 3 P 2 8 + 3 (NH 4 ) 2 C 2 Q 4 = 3 CaC 2 4 + 2 (NH 4 ^PO 4 . 

 2 (NH 4 ) 3 PO 4 + 2 MgSO 4 =2NH 4 MgP0 4 +2(NH^ 2 S0 4 , 

 2 NH 4 MgPO 4 (on ignition)=Mg 2 

 Mg 2 P 2 O7 : Ca 3 P 2 O 8 ::2'5 : percentage 

 (=222) (=310) in i gram of super. 



/. Ca 8 P,0 8 present= 8 ' 5 g X g 319 = 3'49 % 



The total CaCO 3 came partly from CaH 4 P 2 O 8 , partly. ,. r 

 from the Ca 3 P 2 O 8 and the rest from the CaSO 4 present 

 in the super, CaSQ* being due to the H 2 SO 4 which 

 is added to Ca 3 P 2 O8 in the preparation of super. 

 Now, CaH 4 P 2 O 8 + 4NH 4 HO + (NH 4 ) 2 C 2 O 4 . 



=CaC 2 4 +2(NH 4 ) 3 P0 4 . 

 CaC 2 O 4 (on ignition)=CaCO 3 + CO. 

 But CaH 4 P 2 O 8 : CaCO 3 : '.amount of CaH 4 P 2 O 8 present 

 : CaCO 3 which was due to the 



monocalcic phosphate, 



i.*., 234 : ioo::20'o3 : percentage of CaCO 3 due to the 

 monocalcic phosphate. 



/. CaCO 3 due to the CaH 4 P 2 O 8 present= 2 ' 3 x - 



=8'56% 



AAAAA 



