Please show how to find moles of KMnO 4 added, moles of Fe 3+ , mass of Fe 3+ ,

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+

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+

Explanation / Answer

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

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

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