In: Chemistry
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+ |
|||||||||
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