Very rusty with chemistry and looking to review some basics. Came across the following:

Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

What I have so far (guessing, basically):

$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$ [still need to balance but correct right hand side?]

I can't remember the exact rules for forming the right hand side. I just put this in place to match ion charges. What prevents me from writing the right hand side as something else entirely (not sure what but I'm asking, "well, couldn't I figure something out if I break everything apart and re-arrange into another option?").

EDIT: Additional Question Below

Ok, so assume the above reaction was balanced (it's not) and valid. Then, my question is does the reaction go both ways ($\ce{<-->}$)? If so, when/why/how? I'm trying to think back many years ago and it has something to do with "reactivity" and the row of the elements on the periodic table, correct? I'm going to look into it more myself but if someone has a quick "30 second" answer I'm sure it'd be of help to me!

[Assuming this is OK and valid in the forward direction. Again, needs to be balanced.]

$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$

[Is this ever ok/possible?]

$$\ce{KCl + Fe(H2PO4)2 -> FeCl2 + K(H2PO4)}$$

  • $\begingroup$ See, what charge does Fe have? $\endgroup$ Commented Dec 9, 2016 at 20:18
  • 1
    $\begingroup$ @IvanNeretin Since it's iron II, it's charge is 2+, correct? $\endgroup$
    – Jon
    Commented Dec 9, 2016 at 20:53
  • $\begingroup$ @jon $\ce{FeCl}$ if it exists has iron (I) because chloride cannot be (-II). So if you are speaking about iron (II) chloride then it must be $\ce{FeCl_2}$ ;) $\endgroup$
    – ParaH2
    Commented Dec 9, 2016 at 20:58
  • $\begingroup$ @jon Right! Now what is the charge of Cl? $\endgroup$ Commented Dec 9, 2016 at 21:00
  • 1
    $\begingroup$ @IvanNeretin Ah, yes 1-. So it'd have to be FeCl2. Ok, I'll make the edit and update my question accordingly. Thanks! $\endgroup$
    – Jon
    Commented Dec 9, 2016 at 21:06

2 Answers 2


This question was the continuation of your question and since both question are related, I am answering the question by emphasizing the mistake you are doing and adding a relevant portion to explain it.


Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

You need to consider potassium phosphate to be $\ce{K3PO4}$ not $\ce{KH2PO4}$ according to IUPAC's Red book (see sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.)

If $\ce{KH2PO4}$ is meant to be added, the correct term should be potassium dihydrogenphosphate.

Extra info.:

Iron hydrogen phosphate are new in this world and has been recently synthesised and as not as stable as iron phosphates. So, only iron (II and III) phosphate and potassium chloride forms from the reaction.

$$\ce{3FeCl2 + 2K3PO4 ⟶ Fe3(PO4)2 + 6KCl}$$

$$\ce{Fe3(PO4)2 ->[air oxidn⋅,400C] FePO4}$$

or actually

$$\ce{4Fe3(PO4)2 + 3O2 ->[400 C] 8FePO4 + 2Fe2O3}$$

Iron hydrogen phosphate can be made by reacting iron(II) phosphate with dilute hydrochloric acid.

$$\ce{Fe3(PO4)2 + 2HCl → 2FeHPO4 + FeCl2}$$


The molecular formula: $$\ce{ 2 K3PO4 (aq) + 3Ni(NO3)2 (aq) -> Ni3(PO4)2 (s) + 6KNO3 (aq) }$$

The complete ionic formula is: $$ \ce{6K+ (aq) + 2PO4^3- (aq) + 3Ni^2+ (aq) +2NO3- -> Ni3(PO4)2 (s) + 6K^+(aq) + 2NO3^-(aq) }$$

The net ionic formula is: $$ \ce{2PO4^3- (aq) + 3Ni^2+ (aq) -> Ni3(PO4)2 (s) }$$

enter image description here

  • $\begingroup$ Be welcome to ChemSE. Please familiarize yourself mathjax (for mathematical equations) and mhchem (about chemistry like reaction equations). Especially the later will both facilitate (and thus, speed up) editing reaction equations in questions / answers / comments, and offer a nicer representation, too. This may be a starter on both, yet mhchem equally may be used with the LaTeX markup, too (mhchem.github.io/MathJax-mhchem). But don't use it in question titles. $\endgroup$
    – Buttonwood
    Commented Oct 2, 2020 at 11:36
  • $\begingroup$ Why speak of Nickel ? $\endgroup$
    – Maurice
    Commented Jun 1, 2021 at 6:59

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