Question

Calculate the number of milliliters of 0.440 M KOH required to precipitate all of the Fe2+ ions in 187 mL of 0.692 M FeSO4 solution as Fe(OH)2. The equation for the reaction is: FeSO4(aq) + 2KOH(aq) Fe(OH)2(s) + K2SO4(aq)

Answer 1

Answer:

588.2 mL

Explanation:

• FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)

First we calculate how many Fe⁺² moles reacted, using the given concentration and volume of FeSO₄ solution (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):

• moles = molarity * volume

• 187 mL * 0.692 M = 129.404 mmol Fe⁺²

Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:

• 129.404 mmol Fe⁺² * $\frac{2mmolKOH}{1mmolFeSO_4}$ = 258.808 mmol KOH

Finally we calculate the required volume of KOH solution, using the given concentration and the calculated moles:

• volume = moles / molarity

• 258.808 mmol KOH / 0.440 M = 588.2 mL