[janjan]

Hello, I have a problem from analytical chemistry and I just want to check if my answer makes sense and If it could possibly even be correct.

Question:
We have a mixture of two unknown metals and a polymer (Nylon 6 6). We dissolve the mixture in a solution of hydrochloric acid (HCl) and nitric acid (HNO3) - this results in a blue and green color solution, which we then filter.

After adding potassium iodide (KI), brown sediment precipitates in the solution. When amylopectin (gelatinized starch) is added, the solution turns dark blue. And finally If we add Chugaev reagent (1percent solution of Dimethylglyoxime in Ethanol), ping precipitate appears in the solution

Determine the two metals and write all reactions.
(There are also numbers to count the exact amounts of the metals, but I just want to check the reactions

My thinking:
I would say one of the metals is Nickel, since I think during its reaction with HCl green NiCl2 forms:
2 HCl + Ni = NiCl2 + H2
And then the Chugaev reagent makes perfect sense to me:
NiCl2 + 2C4H8O2N2 = (Ni(C4H7O2N2)2) (Nickel dimethylglyoxime) + 2HCl

The second metal could be iron, since:
Fe + 4 HNO3 = Fe(NO3)3 + NO + 2 H2O (Fe(NO3)3 is orange in water, so it is possibly wrong)
Then, If you add KI:
2 Ki + 2 Fe(NO3)3 = 2 Fe(NO3)2 + 2 KNO3 + I2
OR, If you add KI with starch:
Fe3+ + I- = Fe2+ + I2 (I'm not sure about full equation)

That's pretty much what I've got. I will be happy for any help or push toward the goal.
Thanks very much in advance

[AWK]

NiCl2 + 2C4H8O2N2 = (Ni(C4H7O2N2)2) (Nickel dimethylglyoxime) + 2HCl

This reaction is performed with ammonia complex of nickel salt. It is not formed in acidic solution.

Iron(III) nitrate in even small excess of acid is practically colorless.

Iron also react with HCl forming FeCl₂ more easily than nickel NiCl₂. Moreover, FeCl₂ does not react with KI and forms false pink complex with dimethylglyoxime that masks nickel complex.

Think about other metal that does not react with HCl and with nitrate of blue color in solution, even in the excess of acid.

[janjan]

Thanks for your response.

It is starting to make a little more sense, I think
So the second metal could be copper, since:
Cu + HNO₃ -> Cu(NO₃)₂ + NO₂ + H₂O (Cu(NO₃)₂ solution is bluish color)
Then adding KI results in:
I^- + Cu²⁺ -> CuI2 -> CuI + I₂ (CuI is not soluble in water -> I thought it was more less white, so I don't know about the brown precipitate, but we can prove the I₂ by adding starch)

I am also pretty sure, that the second metal is Nickel, since Chugaev reagent turns the solution red / pink. I am not sure how, since, you suggested, that the reaction is performed with ammonia complex of nickel.
Could you please help me slightly on this note.
Thanks very much in advance

[AWK]

In the first post you write down this reaction

NiCl2 + 2C4H8O2N2 = (Ni(C4H7O2N2)2) (Nickel dimethylglyoxime) + 2HCl

Since dimethylglyoxime is a very weak acid (pKa <10) than nickel complex is not formed in the presence of strong acid HCl.
The same concerns strong bases (insolubility of Ni(OH)₂) but blue (not green as nickel salts) soluble ammonia nickel(II) complex will form in any ammonia concentration.

[janjan]

Thank you so much for the answer.

I feel embarrassed, but I found the original paper with the task and I actually forgot to write one sentence 
In my first post, before adding KI, we actually neutralize the solution...