OH= 2.2gx 1mol/40.00gNaOHx 2mol/1molx .025x1L= 2.75x10^-3M
While you are right abut starting with 2.2 g of NaOH to calculate OH
- concentration I have no idea what you are doing here, and the final result is wrong.
Q= (2.5x10^-3)(2.75x10^-3)^2 = 1.8x10^-8
1.8x10^-8x 89.87= 1.6x10^-6g of Fe(OH)2 formed
Is this correct?
Sorry to be blunt, but I don't plan to spend my evening guessing which number means what. I have no idea what you are doing.
Hint: start with finding the answer to B, and think if the final concentration of Fe
2+ is low enough for the precipitation to be (for practical purposes) complete.