Answer:
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