A concentrated solution of ammonia is added to a solution of zinc oxide?

NH3 + ZnI2 => ??

An excess of nitric acid solution is added to a solution of tetraaminecopper (II) sulfate.
HNO3 + ? => ??

Zn forms the Zn(NH3)4^+2 ion. It MAY form some Zn(OH)2 at small amounts of NH3(aq) but I think all will eventually end up as the complex ion.
Cu(NH3)4SO4 + HNO3 ==> neutralizes the NH3.

what would the products be though, not just the NIE

I don't know how to answer the Zn part because ZnO is not soluble in water. So what it the ZnO solution a solution in??
And there will not be a molecular equation if there is no ppt. Anyway, at first, Zn(OH)2 forms and that is soluble in an excess of NH3.
Zn^+2 + 2NH3 + 2H2O ==> Zn(OH)2 + 2NH4^+
then an excess of NH3:
Zn(OH)2 + 4NH3 ==> Zn(NH3)4^+2 + 2OH^-

Cu(NH3)4SO4 + 4HNO3 ==> Cu^+2 + 4NH4^+ + SO4^= + 4NO3^-

Check my work.

Your work is correct! Here is a step-by-step explanation for each reaction.

For the reaction between a concentrated solution of ammonia and zinc oxide:

1. First, we need to recognize that ammonia (NH3) is a weak base that can react with zinc oxide (ZnO), which is a basic oxide.

2. However, since zinc oxide is not soluble in water, we need to assume that the ZnO is dissolved in some other solution or medium.

3. When ammonia reacts with zinc ions (Zn^+2) in solution, it forms a complex ion called Zn(NH3)4^+2. This reaction can be represented as:
Zn^+2 + 4NH3 + 2H2O -> Zn(NH3)4^+2 + 2OH^-

Note that at small amounts of ammonia, some Zn(OH)2 may initially form, but it will eventually convert to the complex ion with excess ammonia.

For the reaction between an excess of nitric acid solution and tetraaminecopper (II) sulfate:

1. Tetraaminecopper (II) sulfate (Cu(NH3)4SO4) is a complex compound containing a copper ion (Cu^+2) coordinated with four ammonia ligands.

2. When nitric acid (HNO3) is added, it acts as a strong acid and can replace the ammonia ligands in the complex with hydrogen ions (H+), leading to protonation.

3. The reaction can be represented as:
Cu(NH3)4SO4 + 4HNO3 -> Cu^+2 + 4NH4^+ + SO4^= + 4NO3^-

In this equation, the tetraaminecopper (II) sulfate breaks down into copper ions (Cu^+2), ammonia ions (NH4^+), sulfate ions (SO4^=), and nitrate ions (NO3^-) due to the reaction with nitric acid.

Please note that the net ion equation you provided already includes the ionic species involved in the reaction.