100 mL of 0.3 M AgNO ₃ solution is mixed with 100 mL of 0.1 M CaCl ₂ solution to precipitate AgCl . If the precipitation reaction occurs with a 75 % yield, what is the final concentration of Ag ^+ ions in the resulting mixture?

Step-by-step solution

First, calculate the initial millimoles of the reacting ions: Millimoles of Ag ^+ from AgNO ₃ = 100 0.3 = 30 mmol Millimoles of Cl ^- from CaCl ₂ = 100 0.1 2 = 20 mmol The precipitation reaction is: Ag ^+ + Cl ^- AgCl (s) From the stoichiometry, 1 mole of Ag ^+ reacts with 1 mole of Cl ^- . Since we have 30 mmol of Ag ^+ and 20 mmol of Cl ^- , Cl ^- is the limiting reagent. Theoretical amount of Ag ^+ that should react = 20 mmol. However, the reaction has a 75 % yield. Therefore, the actual amount of Ag ^+ that rea