mmols FeCl2 = mL x M = ?
mmols KOH = 2x mmols FeCl2
M KOH = mmols KOH/mL KOH. You know mmols kOH and M KOH, solve for mL KOH.
How many milliliters of 0.316 M KOH are needed to react completely with 58.0 mL of 0.297 M FeCl2 solution to precipitate Fe(OH)2? The net ionic equation is:
Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
1 answer
1 answer