Question

1. (4p) In an acidic aqueous solution, Fe2+ ions are oxidized to Fe3+ ions by MnO4": 5Fe2+(aq) + MnO4 (aq) + 8H(aq) → 5Fe3+ (aq) + Mn2(aq) + 4H2O(1) In part A of the experiment, suppose that 1.067 g of Fe(NH4)2(SO4)2.6H2O(s) are placed in a 250 ml Erlenmeyer flask to which 20 mL of water and 8 mL of 3 M H2SO4(aq) are added. The solution was titrated to the end point by adding 26.89 mL of KMnO4(aq) from the buret. a. How many moles of Fe2+ are present initially in the Erlenmeyer flask? b. How many moles of MnO4 are required to reach the endpoint in this titration? C. What is the concentration of the KMnO4 solution in the buret? 2. (4p) In part B of the experiment, suppose it required 23.02 ml of 0.01988 M KMnO4(aq) to reach the end point of a titration of an unknown iron salt (containing Fe2+). This is the same titration reaction as in part A of the experiment. Prior to titration, the sample in the Erlenmeyer flask had 0.585 g of the unknown iron salt, 20 ml of water and 8 mL of 3 M H2SO4(aq). a. How many moles of MnO4 were added in this trial? b. How many moles of Fe2+ must have been in the Erlenmeyer flask? c. What is the mass percent of iron in the unknown iron salt? last question on back of this page:

Answer 1

Q1. (a.) mass Fe(NH₄)₂(SO₄)₂.6H₂O = 1.067 g

moles Fe(NH₄)₂(SO₄)₂.6H₂O = (mass Fe(NH₄)₂(SO₄)₂.6H₂O) / (molar mass Fe(NH₄)₂(SO₄)₂.6H₂O)

moles Fe(NH₄)₂(SO₄)₂.6H₂O = (1.067 g) / (392.14 g/mol)

moles Fe(NH₄)₂(SO₄)₂.6H₂O = 2.72 x 10^-3 mol

moles Fe²⁺ initially present = moles Fe(NH₄)₂(SO₄)₂.6H₂O

moles Fe²⁺ initially present = 2.72 x 10^-3 mol

(b.) moles MnO₄^- required = (moles Fe²⁺ initially present) * (1 mole MnO₄^- / 5 moles Fe²⁺)

moles MnO₄^- required = (2.72 x 10^-3 mol) * (1 / 5)

moles MnO₄^- required = (2.72 x 10^-3 mol) * (0.2)

moles MnO₄^- required = 5.44 x 10^-4 mol

(c.) volume KMnO₄ used = 26.89 mL = 0.02689 L

concentration of KMnO₄ in buret = (moles MnO₄^- required) / (volume KMnO₄ used in Liter)

concentration of KMnO₄ in buret = (5.44 x 10^-4 mol) / (0.02689 L)

concentration of KMnO₄ in buret = 0.02024 M

Q2. (a.) concentration KMnO₄ = 0.01988 M

volume KMnO₄ used = 23.02 mL = 0.02302 L

moles KMnO₄ added = (concentration KMnO₄) * (volume KMnO₄ used in Liter)

moles KMnO₄ added = (0.01988 M) * (0.02302 L)

moles KMnO₄ added = 4.576 x 10^-4 mol

(b.) moles Fe²⁺ = (moles KMnO₄ added) * (5 moles Fe²⁺ / 1 mole MnO₄^-)

moles Fe²⁺ = (4.576 x 10^-4 mol) * (5 / 1)

moles Fe²⁺ = (4.576 x 10^-4 mol) * (5)

moles Fe²⁺ = 2.288 x 10^-3 mol

(c) mass Fe = (moles Fe²⁺) * (molar mass Fe)

mass Fe = (4.576 x 10^-4 mol) * (55.845 g/mol)

mass Fe = 0.128 g

mass percent iron = (mass Fe / mass sample) * 100

mass percent iron = (0.128 g / 0.585 g) * 100

mass percent iron = (0.218) * 100

mass percent iron = 21.8 %