AgCl(s)+KBr(s)→AgBr(s)+KCl(aq)
Moles of silver chloride = 12.0⋅g143.32⋅g⋅mol−1=0.0837⋅mol.
Moles of potassium bromide = 13.0⋅g119.0⋅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⋅mol×187.77⋅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.
