A student determines the % Fe in an iron supplement tablet by redox titration. A 0.5000 g tablet is dissolved in dilute H2SO4, the iron reduced to Fe^2+, and titrated with 0.02000 M KMnO4 until a faint persistent pink color appears (no indicator). 24.50 mL of titrant is required. Balanced reaction: MnO4- + 5 Fe^2+ + 8 H+ -> Mn^2+ + 5 Fe^3+ + 4 H2O.
What is the mass percent of iron in the tablet?
- Acheck_circle
27.4%
- B
54.7%
- C
13.7%
- D
5.47%
Explanation
mol MnO4- = 0.02450 * 0.02000 = 4.90e-4. mol Fe = 5 * 4.90e-4 = 2.45e-3. mass Fe = 2.45e-3 * 55.85 = 0.1368 g. % = 0.1368/0.5000 = 27.4%.