# The no. of moles of KMnO4 that will be needed to react completely with one mole of ferrous oxalate in acidic solution is(a) 3/5  (b)2/5(c)4/5  (d)None of these

A)  3/5......

KMnO4 + FeC2O4 + H ------> Mn2+  + Fe3+  +  2CO2

KMnO4  = 5 electrons

ferrous oxalate in acidic solution = 3 electrons

while reaction exchange of electrons takes place..........so

3 moles of KMnO4 = 5 moles of  FeC2O4

5 moles of  FeC2O4 = 3 moles of KMnO4

so 1 mole of FeC2O4 = 3/5 moles of KMnO4

hence 3/5 moles of KMnO4 of KMnO4 will be needed to react completely with one mole of ferrous oxalate in acidic solution.....

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thanx a ton....

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hey Raghuraman can u give me the full balanced equation if can also the half reactions?

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We know,

For KMnO4,Mn changes it oxidation state from +7 to +2This happens in acidic medium as,It changes from MnO4 to MnSO4 or MnCl2 etcMn(7+) + 5e- gives Mn(2+)Since, per mole of the compound, 5 electrons are lost ,Valence factor is 5.For ferrous oxalate,Fe(2+) gives Fe(3+) + e-AndC2O4 (2-) gives 2CO2 + 2e-We find that here, per molecule of ferrous oxalate, 3e- are lost.So valence factor is 3Therefore,Equivalents of potassium permanganate= Equivalents of ferrous oxalateEquivalents= moles x valence factor.Moles of potassium permanganate =( 3 x 1)/ 5 =3/5I hope you all understand this. If any doubts, please ask. If any mistakes, please forgive me and point the out.... Thankyou
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Lengu stika - dodo ah u gotta Answeraa. Notnja
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Option A
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Option A.....
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A)3/5
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option A
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😁😁
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A) ?3/5......

KMnO4 + FeC2O4 + H ------> Mn2+ ?+ Fe3+ ?+ ?2CO2

KMnO4 ?= 5 electrons

ferrous oxalate in acidic solution = 3 electrons

while reaction exchange of electrons takes place..........so

3 moles of KMnO4 = 5 moles of ?FeC2O4

5 moles of ?FeC2O4 =?3 moles of KMnO4

so 1 mole of?FeC2O4 = 3/5?moles of KMnO4

hence 3/5?moles of KMnO4 of?KMnO4?will be needed to react completely with one mole of ferrous oxalate in acidic solution.....
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I could not find dimentional analysis here..pls explain how to study that chapter
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3KMnO4 + 5FeC2O4 +24H+ ?5Fe3 + +10CO2 +3Mn2+ +3K+ +12H2O

By Unitary method

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Here is the solution of the question asked by you-

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CH5N2O
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KMnO4 + FeC2O4 + H ------> Mn2+ ?+ Fe3+ ?+ ?2CO2

KMnO4 ?= 5 electrons

ferrous oxalate in acidic solution = 3 electrons

while reaction exchange of electrons takes place..........so

3 moles of KMnO4 = 5 moles of ?FeC2O4

5 moles of ?FeC2O4 =?3 moles of KMnO4

so 1 mole of?FeC2O4 = 3/5?moles of KMnO4

hence 3/5?moles of KMnO4 of?KMnO4?will be needed to react completely with one mole of ferrous oxalate in acidic solution.....
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Ans. B. 2/5
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D)none of these

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Help

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lg cl2(g)
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None of these
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Vemagak
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sorry i
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Hi hello
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If 6.3of sodium bicarbonate are added to 15g of CH3COOH solution, the residue is found to weight 18.0g.what is the mass of CO2 released
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A) ?3/5......

KMnO4 + FeC2O4 + H ------> Mn2+ ?+ Fe3+ ?+ ?2CO2

KMnO4 ?= 5 electrons

ferrous oxalate in acidic solution = 3 electrons

while reaction exchange of electrons takes place..........so

3 moles of KMnO4 = 5 moles of ?FeC2O4

5 moles of ?FeC2O4 =?3 moles of KMnO4

so 1 mole of?FeC2O4 = 3/5?moles of KMnO4

hence 3/5?moles of KMnO4 of?KMnO4?will be needed to react completely with one mole of ferrous oxalate in acidic solution.....
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??9o
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