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## Q. 4.10

Consider the balanced equation for the reaction between iron(II) and permanganate ions in acidic solution:

$MnO^{-}_{4}(aq) + 5Fe^{2+}(aq) + 8H_{+}(aq) → 5Fe^{3+}(aq) + Mn^{2+}(aq) + 4H_{2}O$

What 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

## Verified Solution

 Fe$(NO_{3})_{2}$ (parent) → $Fe^{2+}$ (ion) K$MnO_{4}$ (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 K$MnO_{4}$ moles = V × M; V = $\frac{0.00138 mol}{0.684 mol/L}$ = 0.00202 L = 2.02 mL 6. V K$MnO_{4}$