1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 mol−1, NaH2PO4=120g mol−1) NaH2PO4+Mg2++NH+4→Mg(NH4)PO4.6H2O→Mg2P2O7

A
2.1 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3.0 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5.0 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.2 g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 1.2 gNaH2PO4+Mg2++NH+4→Mg(NH4)PO4.6H2O→Mg2P2O7 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

Suggest Corrections
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program