Question

2.5g of AR ammonium iron (ll) sulfate hexahydrate crystals are dissolved in 40mL of 1M sulfuric acid solution. Solution is then titrated with 6.4mL of 0.02M potassium permanganate (VII) solution.

a) Write an overall equation for the reaction.

b) How many moles of permanganate are present in the titre value?

c) How many moles of Fe^2+ are present in the solution?

d) What mass of Fe^2+ is present in the ammonium iron (ll) sulfate hexahydrate?

e) What is the percentage of Fe^2+ in ammonium iron (ll) sulfate hexahydrate?

Answer 1

Hi Dear Friend, i assume you are doing online homework and it requires correct number of significant figures answer. Hence in final answer i tried to maintain 2 sig fig, if you want more exact just use immediate above line end number that is also bold.

a) Write an overall equation for the reaction.

Overall net ionic equation can be written as

5 X (Fe2 Fe3+ + e') MnO48H5e° -Mn2+ + 4H20 oxidation reduction MnO45Fe2+ 8H* -> Mn2+ + 5Fe3+ + 4H20 overall process

Answer: 2 KMnO₄ + 8 H₂SO₄ + 10 FeSO₄ ⟶ K₂SO₄ + 2 MnSO₄ + 8 H₂O + 5 Fe₂(SO₄)₃

b) How many moles of permanganate are present in the titre value?

6.4 mL x 10^-3 L x 0.02 M = 0.000128 moles

Answer: 0.00013 moles

c) How many moles of Fe²⁺ are present in the solution?

Ammonium iron (ll) sulfate hexahydrate, also called as Mohr's salt generally represented as [ (NH4)2Fe(SO4)2.6H2O ] or simply [ FeH₂₀N₂O₁₄S₂ ] and it's molar mass = 392.1 g/mol

moles Fe²⁺ = 2.5 g / 392.1 g/mol = 0.0063759245 moles

Answer: 0.0064 moles

d) What mass of Fe²⁺ is present in the ammonium iron (ll) sulfate hexahydrate?

mass of Fe²⁺ = moles Fe²⁺ x molar mass of Fe²⁺ = 0.0063759245 mol x 55.845 g/mol = 0.3560635042 g

Answer: mass of Fe²⁺ is present in the ammonium iron (ll) sulfate hexahydrate = 0.36 g

e) What is the percentage of Fe²⁺ in ammonium iron (ll) sulfate hexahydrate?

percentage of Fe²⁺ = [mass of Fe²⁺ / mass of AR ammonium iron (ll) sulfate hexahydrate ] x 100%

percentage of Fe²⁺ = [0.3560635042 g / 2.5 g] x 100% = 14.2425401683 %

Answer: the percentage of Fe²⁺ in ammonium iron (ll) sulfate hexahydrate = 14 %

Hope this helped you!

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