Q, What volume of 0.472 mol/L AgNO3 will precipitate the chloride ion in 40ml of 0.183 mol/L AlCl3?

Question

- Answer is 46.6ml, but how to solve this question?

Answer 1

Moles Ag= Volume*Molarity
But moles of Ag has to equal moles of Cl ion (it forms AgCl).
Moles Cl=.04*.184*3 (three ions in AlCl3)
set them equal, and solve for Volume