If I mix 25.0mL of 3.00 M Lead (II) Nitrate with 45.0mL of 2.50 M Potassium Chloride.
What is the final number of moles of potassium ion present? What is the final concentration of potassium ion? What is the concentration of the excess ion at the end of the reaction?
I know the equation is
Pb(NO3)2 + 2KCl --> PbCl2 + 2KNO3