Write the balanced net ionic equation for the reaction between MnO_4^- ion and F

Question

Write the balanced net ionic equation for the reaction between MnO_4^- ion and Fe^2+ ion in acid solution. How many moles of Fe^2+ ion can be oxidized by 1.3 times 10^-2 moles MnO_4^2- ion in the reaction in Question 1? moles A solid sample containing some Fe^2+ ion weighs 1.264 g. It requires 38.67 mL 0.02487 M KMnO_4 to titrate the Fe^2+ in the dissolved sample to a pink end point. How many moles of MnO_4^- ion are required? How many moles of Fe^2+ are there in the sample? How many grams of iron are there in the sample? What is the mass percent of Fe in the sample? What is the mass percent of Fe in iron(II) ammonium sulfate hexahydrate, Fe(NH_4)_2 (SO_4)_2 middot 6 H_2O?

Explanation / Answer

1. Fe2+ + 1e– Fe3+ 5[Fe2+ + 1e– Fe3+ ] 5 Fe2+ + 5 e– 5 Fe 3+

MnO 4 – Mn2+ + 5e– (balance the 1/2 reaction)2 8H+ + MnO 4 – Mn2+ + 4H2O + 5e– Balanced redox reaction :

8H+ + MnO 4 – + 5Fe3+ 5Fe2+ + Mn2+ + 4H2O

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Q2

=0.013 mol MnO 4 – x (5 Fe3+ /1 MnO4 ) = 0.065 mol Fe3+ answer –

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Q3

A solid sample containing some Fe2+ ion weighs 1.264 g. It requires 38.67 mL 0.02487 M KMnO4 to
titrate the Fe2+ in the dissolved sample to a pink end point.

a)
(0.03867 L) x (0.02487 mol/L KMnO4) x (1 mol MnO4{-} / 1 mol KMnO4) = 0.0009617 mol MnO4{-}

b)
MnO4{-} + 5 Fe{2+} + 8 H{+} Mn{2+} + 5 Fe{3+} + 4 H2O

(0.0009617 mol MnO4{-}) x (5 mol Fe{2+} / 1 mol MnO4{-}) = 0.0048085 mol Fe{2+}

c)
(0.0048085 mol Fe{2+}) x (55.8452 g Fe/mol) = 0.2685 g Fe

d)
(0.2685 g Fe) / (1.264 g total) = 0.2124 = 21.24% Fe by mass

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