[capp--city]

I have been tasked with finding the net ionic equation of the identification process of the cation Al⁺³. My steps were as follows:

1.) Place an eyedropper full of aluminum solution in a test tube.
2.) Add one eyedropper full of HCl to make the solution acidic.
3.) Add one dropper of NH₃ to make the mixture basic.
4.) A clear white precipitate indicates the presence of aluminum.

So far I have the equation for the first reaction:

2Al(s) + 6HCl(aq) ---> 3H₂(g) + 2AlCl₃(aq)

The second reaction is where I'm confused. I assume the reactants would be this:

3H₂(g) + 2AlCl₃(aq) + NH₃(aq) ---> ?

I'm not sure what would form from those reactants. The next step would be to put the second reaction into the net ionic equation, but I cannot do that yet since I don't know what the products would be. Any help is appreciated, thanks!

 

[Xenonman]

Do you know of an aluminum white precipitate? Knowing what you are shooting for should help you getting there.
Also, dihydrogen is a gas. I suppose most would escape from the solution, making no reaction you are interested in right now.
 

[Borek]

1.) Place an eyedropper full of aluminum solution in a test tube.

What does it mean "aluminum solution"? Is it suspension of metallic Al, or solution of some salt? If the latter, reaction

2Al(s) + 6HCl(aq) ---> 3H₂(g) + 2AlCl₃(aq)

doesn't make much sense. If the former, it is not a solution, but suspension.

2.) Add one eyedropper full of HCl to make the solution acidic.
3.) Add one dropper of NH₃ to make the mixture basic.
4.) A clear white precipitate indicates the presence of aluminum.

If I remember correctly, this procedure can also mean presence of at least two other metals.