I figured out the NO3- which is 0.13, I'm stuck on the Cl- I have 0.24*2-0.0416= 0.4384 but I'm being told that's wrong
What mass of silver chloride can be prepared by the reaction of 160.0 mL of 0.26 M silver nitrate with 150.0 mL of 0.24 M calcium chloride?
Cl ‾
NO3‾
Ca2+
You have 0.26M*160 mL = 41.6 millimols Ag^+.
You have 0.24M x 150 mL x 2 = 72.0 mmols Cl^-.
So you have 72.0-41.6 = 30.4 mmols Cl^- in excess.
(Cl^-) = mmols/mL = ? about 0.098M
