[kverge]

The question is: Aqueous Chlorine is added to a solution of aluminum bromide in a single replacement reaction which forms to precipitate and bromine liquid.

The equation I got is: 3Cl2(aq)+2AlBr3(aq) --> 3Br2(l)+2AlCl3

What I don't know is if the last part (2AlCl3) is a solid or an aqueous solution. I need to know in order to write an ionic equation and I have no idea how to tell.

Thanks.

[Hunter2]

Everythingis in liquid.

The basic reaction is Cl₂ + 2 Br^- => Br₂ + 2 Cl^-

Aluminium is not affected and stays as Al³⁺. 

What can happen is that a Bromo or chloro or mixed complex is formed.  [AlCl_nBr_m]^-, n = 0 to 4, m = 4-n

[Borek]

Everythingis in liquid.

Which is a very misleading way of answering the question, as instead of helping to differentiate between (l) and (aq) it only confuses things further.

What I don't know is if the last part (2AlCl3) is a solid or an aqueous solution.

You started with two water solutions, you have enough water to keep all ionic substances dissolved (unless they are weakly soluble, but AlCl₃ is well soluble). Bromine is a liquid. Everything else is in the form of water solution, so (aq).

Whether the Br₂ will drop out as a liquid from the water solution depends on the concentrations of all substances present, it is not guaranteed you will see liquid bromine here (actually I am not convinced you will see it at all).

[kverge]

Thank you for the *delete me*