Here is the answer for the question – **If you collected a 3.02g sample of MgNH4PO4x6H2O calculate a the value for the mass of P2O5 in the initial mass of fertilizer analyzed.**. You’ll find the correct answer below

### If you collected a 3.02g sample of MgNH4PO4x6H2O calculate a the value for the mass of P2O5 in the initial mass of fertilizer analyzed.

What is the % P2O5 in the fertilizer?

**The Correct Answer is**

**3.02gMgNH4PO4x6H2O (1mol/245.41g) x(1mol P2O5/2mol)x 141.94gP2O5/1mol=0.87335g P2O50.87335g/3.06g x100=28.54%**

### Reason Explained

3.02gMgNH4PO4x6H2O (1mol/245.41g) x(1mol P2O5/2mol)x 141.94gP2O5/1mol=0.87335g P2O50.87335g/3.06g x100=28.54% is correct for If you collected a 3.02g sample of MgNH4PO4x6H2O calculate a the value for the mass of P2O5 in the initial mass of fertilizer analyzed.

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