#AgCl(s) + KBr(s) rarr AgBr(s) + KCl(aq)#
#"Moles of silver chloride"# #=# #(12.0*g)/(143.32*g*mol^-1)=0.0837*mol.#
#"Moles of potassium bromide"# #=# #(13.0*g)/(119.0*g*mol^-1)=0.109*mol.#
Clearly, there is sufficient bromide anion to effect metathesis, and should the reaction go to completion, then #0.0837*mol# #KBr# would eventually precipitate, which constitutes a mass of #0.0837*molxx187.77*g*mol^-1=15.72*g#. I think you can calculate the mass of the excess potassium bromide.
Had this reaction been performed, you would see the white precipitate of #AgCl# change to the cream-coloured #AgBr#. While silver halides are both quite insoluble, #AgBr# is more insoluble than #AgCl#, and this would drive the reaction to the right as written. The problem with this reaction is that both silver salts are photo-active, and would reduce to give metallic silver as a dark precipitate.
