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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}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

fig 4.12
fig 4.6

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}