*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?
2 answers
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)
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)
2 answers