Question

How many milliliters of 0.20 m alcl3 solution would be necessary to precipitate all the ag+ form 45 ml of 0.20m agno3?

Answer 1

Answer:

45 ml x (1L / 1000 ml) x 0.2 (moles/L) = 0.009 moles AgNO3

0.009 moles AgNO3 x (1/3) / (0.20 moles/L) x (1000 ml/L) = 15 ml 0.20 M AlCl3