Chapter 4
Q. 4.10
Consider the balanced equation for the reaction between iron(II) and permanganate ions in acidic solution:
MnO^{-}_{4}(aq) + 5Fe^{2+}(aq) + 8H_{+}(aq) → 5Fe^{3+}(aq) + Mn^{2+}(aq) + 4H_{2}OWhat volume of 0.684 M KMnO_{4} solution is required to completely react with 27.50 mL of 0.250 M Fe(NO_{3})_{2} (Figure 4.12)?
| ANALYSIS | |
| V (27.50 mL) and M (0.250) of Fe(NO_{3})_{2} M (0.684) of KMnO_{4} |
Information given: |
| reacting species; stoichiometric ratios | Information implied: |
| volume of KMnO_{4} | Asked for: |
STRATEGY
Follow the flow chart shown in Figure 4.6.
V × M → mol parent : mol ion → mol ion → mol parent → V × M
Step-by-Step
Verified Solution
| Fe(NO_{3})_{2} (parent) → Fe^{2+} (ion) KMnO_{4} (parent) → MnO_{4}^{-} (ion) |
1. Parent → ion |
| V × M = (0.02750 L)(0.250 mol/L) = 0.00688 | 2. mol Fe(NO_{3})_{2} |
| 0.00688 mol Fe(NO_{3})_{2} × \frac{1 mol Fe^{2+}}{1 mol Fe(NO_{3})_{2}} = 0.00688 | 3. mol Fe^{2+} |
| 0.00688 mol Fe^{2+} × \frac{1 mol MnO_{4}^{-}}{5 mol Fe^{2+}} = 0.00138 | 4. mol MnO_{4}^{-} |
| 0.00138 mol MnO_{4}^{-} × \frac{1 mol KMnO_{4}}{1 mol MnO_{4}^{-}} = 0.00138 | 5. mol KMnO_{4} |
| moles = V × M; V = \frac{0.00138 mol}{0.684 mol/L} = 0.00202 L = 2.02 mL | 6. V KMnO_{4} |