What mass of KCl is needed to precipitate the silver ions from 23.0mL of 0.110 M AgNO3 solution

Hey guys. I was working through my homework, and came across this problem. I've found the moles/liter of the solution (.00253). I think I've balanced the chemical equation KCl + AgNO3 --> KNO3 (aq) + AgCl (s), but am unsure of how to proceed.

Any help would be much appreciated!

moles/liter of the solution = 0.11M AgNO3, not 0.00253

KCl + AgNO3 --> KNO3 + AgCl
0.023L x 0.11M AgNO3 = 0.00253moles AgNO3
this requires 0.00253moles KCl
0.00253moles KCl x 74.55g/mole = 0.189g required