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Question

If 1.11 g Mg2P2O7 is formed at the end of the reaction, then what weight of NaH2PO4 was present originally given that no other products contained phosphorus?
(Molar mass Mg2P2O7=222 g mol1, NaH2PO4=120g mol1)
NaH2PO4+Mg2++NH+4Mg(NH4)PO4.6H2OMg2P2O7 
 
  1. 5.0 g
  2. 1.2 g
  3. 3.0 g
  4. 2.1 g


Solution

The correct option is B 1.2 g
NaH2PO4+Mg2++NH+4Mg(NH4)PO4.6H2OMg2P2O7 
Applying POAC on P,
Moles of P in NaH2PO4= moles of P in Mg2P2O7  
1 × number of mol of NaH2PO4=2 ×number of mol of Mg2P2O7   
Weight of NaH2PO4molar mass=2×Weight of Mg2P2O7molar massWeight of NaH2PO4120=2×1.11222  
Weight of NaH2PO4  =1.2 g
 Thus, the weight of NaH2PO4 present originally was 1.2 g
 

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