A450.0 ml sample of a 0.242 m solution of silver nitrate is mixed with 400.0 ml of 0.200 m calcium chloride. what is the concentration of cl- in solution after the reaction is complete?

0.538 M of Cl- in solution

Explanation:

First we need the balanced reaction:

2AgNO3 + CaCl2  ⇒  Ca(NO3)2 + 2AgCl

Then , we calculate mol of Ag in the reaction.

mol Ag=  450mL * (0.242 molAgNO3/ 1000mL) = 0.1089 mol AgNO3

Now, how many moles of CaCl2 we need to react with 0.1089 mol AgNO3? We take this information, from balanced reaction.  

0.1089 mol AgNO3* (1 mol CaCl2/ 2 mol AgNO3) = 0.05445 mol CaCl2

We just need 0.054 mol CaCl2,  and the rest of CaCl2 remain in solution.

so,

0.08 mol CaCl2 - 0.05445 mol CaCl2 = 0.0256 mol CaCl2 in excess,

Now, how many moles of Cl- are in 0.0256 mol of CaCl2?

0.0256 mol CaCl2* 2 mol Cl-/1mol CaCl2= 0.0511 mol Cl-

Molarity Cl- = Moles Cl-/ Volumen (L)

V (L) = (450 mL + 400 mL) * (1L/ 1000 mL) = 0.950 L

M = 0.0511mol Cl-/ 0.950 L = 0.538 M

it's filtration, but this is an assumption so my apologies if i got this