Need some help with chemistry please.

What mass of silver chloride can be prepared by the reaction of 180.0 mL of 0.26 M silver nitrate with 190.0 mL of 0.17 M calcium chloride?
g=6.7
Calculate the concentrations of each ion remaining in solution after precipitation is complete. (Enter a 0 if none of the ion remains.)

Ag+ 0 M
Cl ‾ ?
NO3‾ .13 M
Ca2+ .087M


Okay so I'm totally lost on getting Cl and i only have one attempt left. please show me how to do it (I changed the numbers around a bit so it won't do me much good).

The number of moles of Cl- that were initially added will reduce by the number of moles that form AgCl with the Cl- ions

Ag+ + Cl- ----> AgCl(s)

moles Ag+ ions = 0.26 M x 0.1800 L = 0.0468 moles
Ag+ reacts with Cl- in a 1:1 ratio.
Therefore number moles Cl- that are used = 0.0468 mol

initial moles Cl-, don't forget that each CaCl2 contains 2 Cl- ions
moles Cl- = = 2 x 0.17 M x 0.1900 L = 0.0646 mol

moles Cl- remaining = 0.0646 - 0.0468 mol = 0.0178 mol

Molarity Cl- = 0.0178 mol / 0.370 M = 0.048 M