How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?

FeCl3 + 3NH4OH Fe (OH) 3 + 3NH4Cl

9.9 ml of 0.200M NH₄OH (aq)

Explanation:

3NH₄OH (Iaq) + FeCl₃ (aq) = > NH₄Cl (aq) + Fe (OH) ₃ (s)

? ml of 0.200M NH₄OH (aq) reacts completely with 12ml of 0.550M FeCl₃ (aq)

1 x Molarity NH₄OH x Volume Am-OH Solution (L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution

1 (0.200M) (Vol Am-OH Soln) = 3 (0.550M) (0.012L)

=> Vol Am-OH Soln = 3 (0.550M) (0.012L) / 1 (0.200M) = 0.0099 Liter = 9.9 milliliters