So, this is going to be more like 2 or 3 questions. Here goes.

First, what is the net-ionic reaction for Fe (III) + K4Fe(CN)6 be? I've been agonizing over this for HOURS with no answer in sight. (I don't even know what to put down as a product!)

Second, what would the net-ionic reaction be for Cu2+ + NH4OH --> [Cu(NH3)4]?

Third, if trying to find chloride in an unknown, would it be sensible to use AgCl? (I think this is what we actually did use when we ran the lab in class) and how would I show a net-ionic reaction for that? Would it be Ag+ +Cl- --> AgCl (I think I may or may not have been over complicating things)

and lastly, how would I run a test for Nitrate? That isn't the brown ring test?

I'm so sorry this is so very long! I'd really greatly appreciate any and all help given!

First: Prussian blue is the compound formed. The ideal formula is Fe4[Fe(CN)6]3 and I usually call this ferric ferrocyanide but the new IUPAC name would be iron(III) hexacyanoferrate(II).

2. I would write it as
Cu^+2 + 4NH3(aq) ==> Cu(NH3)4^+2(aq).
Where did the NH4OH go? Technically,
NH3 + HOH ==> NH4^+ + OH^- but that is a weak base and not many of the ions are present. Aqueous NH3, which I wrote above as NH3(aq), is NH3 in water solution. Years ago, and I learned it this way too, we assumed NH3 reacted with H2O to form NH4OH and that dissociated into NH4^+ and OH^-. While it is true that the NH4^+ and OH^- ions are formed, it has been concluded that the compound, NH4OH, does not exist. So we write the equilibrium as
NH3 + HOH ==> NH4^+ + OH^- and we do not show the intermediate of NH4OH.
In fact, we write the Kb for NH3 as
(NH4^+)(OH^-)/(NH3)

Third question. When trying to determine the amount of chloride in an unknown, yes, one would use the AgNO3 addition to form AgCl. The net ionic equation is as you wrote it. And yes, the brown ring test is the nitrate test.

I see. Thanks so much for your help!

No problem at all! I'll try my best to help you with your questions.

1. The net-ionic reaction for Fe (III) + K4Fe(CN)6 is determined by breaking down the compound into its ions, removing any spectator ions, and writing the balanced equation for the remaining ions. Let's break it down step by step:

Fe (III) + K4Fe(CN)6 = Fe^3+ + 4K+ + [Fe(CN)6]4-

Now, we need to identify the spectator ions (ions that do not take part in the reaction): K+ ions are present on both sides of the equation, so they are spectator ions.

The net-ionic equation includes only the ions involved in the actual reaction. In this case, remove the spectator ions (K+):

Fe^3+ + [Fe(CN)6]4- --> Fe[Fe(CN)6]3

The product is Fe[Fe(CN)6]3.

2. To determine the net-ionic reaction for Cu2+ + NH4OH --> [Cu(NH3)4], follow the same steps as before:

Cu2+ + NH4OH = Cu^2+ + NH4+ + OH-

The spectator ions are NH4+ and OH-, which are present on both sides of the equation. Remove them:

Cu^2+ + OH- --> [Cu(NH3)4]^2+

The product is [Cu(NH3)4]^2+.

3. When trying to find chloride in an unknown, it is sensible to use AgCl because silver chloride is insoluble in water and forms a white precipitate. The net-ionic reaction would be:

Ag+ + Cl- --> AgCl

The product is AgCl, which appears as a white solid (precipitate).

4. To test for nitrate (NO3-), the brown ring test is one method, but if you're looking for an alternative, you can try the nitrate ion test using a mixture of nitrate and sulfamic acid. Here's how you can perform the test:

a. Take a small amount of the unknown compound.
b. Add a few crystals of sulfamic acid to the test tube.
c. Heat the mixture gently.
d. If a gas with a choking smell (nitrogen dioxide) is evolved, it indicates the presence of nitrate ions.

Remember, it's always essential to exercise caution and follow proper lab safety protocols when conducting experiments or tests.

I hope this helps! Let me know if you have any further questions or need additional explanations.