Question

Please show how to find moles of KMnO₄ added, moles of Fe³⁺, mass of Fe³⁺, and mass percent of Fe³⁺

For my standardization of KMnO₄ the average molarity was 2.11x10^-2M I'm not sure if that's needed to solve

a) Balanced net ionic equation for the redox reaction of Fe²⁺and MnO₄^-: 5Fe⁺²+ MnO₄^-2+ 8H⁺Þ 5Fe⁺³+Mn⁺²+ 4H₂O

b) Data:

Run	Mass of Green Salt (g)	Vol. KMnO4 (mLs)	– Vol. blank (mLs)	Difference in vol. (mLs)		Moles KMnO4 added	Moles Fe3+	Mass of Fe3+ (g)	Mass % Fe3+
1	.1168g	4mL	.2mL	3.8mL
2	.1038g	3.1mL	.2mL	2.9mL
					Average mass % Fe3+

Answer 1

b) Start with the balanced chemical equation.

5 Fe²⁺ + MnO₄^- + 8 H⁺ -----> 5 Fe³⁺ + Mn²⁺ + 4 H₂O

As per the balanced chemical equation,

5 moles Fe³⁺ = 1 mole MnO₄^-

Fill up the table as given.

Run	1	2
a) Mass of green salt (g)	0.1168	0.1038
b) Vol. of KMnO4 (mL)	4.0	3.1
c) Vol. of blank (mL)	0.2	0.2
d) Difference in vol (mL) (b – c)	3.8	2.9
e) Moles KMnO4 added = (d/1000)*(2.11*10-2 M) (check sample calculation I)	8.018*10-5	6.119*10-5
f) Moles Fe3+ (check sample calculation II)	4.009*10-4	2.4476*10-4
g) Mass Fe3+ (g) (check sample calculation III)	0.02239	0.01367
h) Mass % Fe3+ (check sample calculation IV)	19.1695	12.9206
i) Average mass % Fe3+ (check sample calculation V)	16.04505 ≈ 16.045

Sample calculation I:

Moles of KMnO₄ taken = (vol. of KMnO₄ in L)*(concentration of KMnO₄ in mol/L) = (3.8 mL)*(1 L/1000 mL)*(2.11*10^-2 mol/L) = 8.018*10^-5 mol.

Sample calculation II:

As per stoichiometric equation above, 1 mole MnO₄^- = 5 moles Fe³⁺.

Therefore, 8.018*10^-5 mole KMnO₄ = (8.018*10^-5 mole MnO₄^-)*(5 moles Fe³⁺/1 mole MnO₄^-) = 4.009*10^-4 mole.

Sample calculation III:

Molar mass of Fe = 55.85 g/mol.

Therefore, mass of Fe³⁺ = (4.009*10^-4 mole)*(55.85 g/mol) = 0.02239 g

Sample calculation IV:

Mass% Fe³⁺ = (mass of Fe³⁺/mass of sample)*100 = (0.02239 g/0.1168 g)*100 = 19.1695

Sample calculation V:

Average mass% Fe³⁺ = (19.1695 + 12.9206)/2 = 16.04505