Calculate the number of millilitres of NH3(aq) solution (d = 0.986 g/ mL) contain 2.5 % by weight NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8 g sample that contains 50 % Fe2O3.

Answer 1

wt Fe2O3 = 0.8 x 1/2 = 0.4gram.
Convert 0.4g Fe2O3 to g Fe. That's
0.4g x (2 atomic mass Fe/molar mass Fe2O3) = estimated 0.28g Fe = estd 0.005 mol Fe (that's mol = g/atomic mass).

So how much OH^- is needed?
Fe^3+ + 3OH^- ==> Fe(OH)3
0.005 mols Fe^3+ will require 3*0.005 mols OH^- = about 0.015

What's the concentration of the NH3? That's
1000 mL x 0.0986 g/mL x 0.025 x (1/molar mass NH3) = estd 0.145M
Then M NH3 = mols NH3/L NH3. You have M and you have mols NH3 needed, solve for L NH3 needed and convert to mL.