*What are the products after a single replacement rxn?

*Determine the spectator ion(s) and the net ionic equation.

F2(g) + AlBr3(aq)--->?

After single replacement....Br2 is a liquid, not a solid so ppt will not form..hence, there is no reaction?

Please don't erase double posts until you KNOW you can without a problem. This is the second or third time I've posted an answer to a double post and the answer and post disappears. Then I must do all of that typing over. Thanks.

Part of what you've said is true; part is not true. There is a reaction and Br2 is a liquid.

3F2(g) + 2AlBr3(aq)---> 2AlF3(aq) + 3Br2(l)

Now you write the full ionic equation which I will do but you put in the phases.
3F2 + 2Al^3+ + 6Br^- ==> 2Al^3+ + 6F^- + 3Br2

Now cancel those items on each side that are common. I see 2Al^3_ on each side so we can cancel them. That leaves us with the net ionic equation of
3F2(g) + 6Br^-(aq) ==> 3Br2(l) + 6F^-(aq)

In order to determine the products of a single replacement reaction, we need to identify the activity series of the elements involved. The activity series is a list that ranks metals in order of their reactivity. If an element is higher in the activity series than the other element, it can displace that element from its compound.

In this reaction, we have F2, a nonmetal, reacting with AlBr3, a compound containing the metal aluminum. To determine if a reaction occurs, we need to compare the reactivity of these two elements.

The activity series tells us that fluorine (F2) is more reactive than aluminum (Al). Therefore, F2 can displace aluminum from its compound, AlBr3. The balanced chemical equation for this single replacement reaction is:

2F2(g) + 2AlBr3(aq) -> 2AlF3(s) + 3Br2(l)

So in this reaction, the products are aluminum fluoride (AlF3) and liquid bromine (Br2).

Now, to determine the spectator ions and write the net ionic equation, we need to first write out the complete ionic equation.

2F2(g) + 6NaI(aq) + 6Cl-(aq) -> 2Na+(aq) + 2AlF3(s) + 6I-(aq) + 3Br2(l)

Next, we identify the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the reaction. In this case, the spectator ions are Na+ and I-.

To write the net ionic equation, we remove the spectator ions from the equation:

2F2(g) + 2AlBr3(aq) -> 2AlF3(s) + 3Br2(l)

This is the net ionic equation for the reaction, which shows only the species that participate in the reaction.