February 24, 2024, 03:42:05 AM
Forum Rules: Read This Before Posting

Topic: Equivalence Mass  (Read 6310 times)

0 Members and 1 Guest are viewing this topic.

Offline Sturm88

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Equivalence Mass
« on: September 13, 2008, 11:14:17 PM »
Phosphorous in a Fertilizer

"P205" (cited by manufacturer) --> (NH4)H2PO4 (actual fertilizer) --> MgNH4PO4.6H2O (product)

So P2O5 apparently isn't the actual phosphorous in fertilizer it's ammonium dihydrogen phosphate.

If I have 1 gram of the fertilizer (MgNH4PO4.6H20) initially what's the equivalence mass of the imaginary "P2O5"

I've thought about it and I'm a little confused. Not sure what to use.

Since there are two hydrogens in the actual fertilizer and 2P's in the phosporic acid... I have tried this.

MolarMass of P2O5 / 2 for equivalence mass... but I'm not sure if that's what is needed to be done.

Online Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27600
  • Mole Snacks: +1794/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Equivalence Mass
« Reply #1 on: September 14, 2008, 05:32:37 AM »
Try to write reaction of fertilizer "synthesis" that starts with P2O5. Something like CaO + P2O5, where there is obviously 141.9 g of P2O5 in 321.2 g of Ca3(PO4)2.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links