Question

How many milliliters of 0.20 MAICI solution would be necessary to precipitate al
the Ag* form 45 mL of a 0.20 M AgNO3 solution?
AICI, (aq) + 3AgNO (aq) -> Al (NO3)3 (aq)
(a) 15 ml
(b) 30 mL
(c) 45 mL
(d) 60 ml
AM​

Answer 1

Answer:

option b is the correct answer

Answer 2

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