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1 mol ferric oxalate is oxidized by x mol MnO4- and 1 mol ferrous oxalate is oxidized by y mol of MnO4- in acidic medium. Thr ratio x/y is ?

1 mol ferric oxalate is oxidized by x mol MnO4- and 1 mol ferrous oxalate is oxidized by y mol of MnO4- in acidic medium. Thr ratio x/y is ?

Grade:11

2 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
From the chemical Reaction:
Fe2(C2O4)3 + MnO4 ===========>>>   Mn2+     +     CO2 
The n factor of Fe2(C2O4)3 would thus be able to be taken as equivalent to the n factor of MnO4- 
Along these lines 
Reactant (n×nf) = Product (n×nf) 
1×6= x×5 
Along these lines x = (6/5) 
Presently on account of ferrous oxalate Fe(C2O4) 
Presently the new response can be composed as: 
MnO4-+ FeC2O4 - Mn2+ +Fe3+ +CO2 
Here additionally Mn is getting diminished from +7 oxidation state to +2 oxidation state. 
Additionally in the compound, ferrous oxalate, Fe is in +2 oxidation state and C is in +3 oxidation state, which are getting changed to +3 and +4 resp. 
So the aggregate n factor for the compound is 1+2=3 
Again keeping the counterparts of the reactants and the items to be same on both the sides, 
Y×5=1×3 
Subsequently y = (3/5) 
So the proportion, (x/y) will be (6×5)/(5×3) 
= 2:1 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to your problem.
 
The reaction taking place is
Fe2(C2O4)3 + MnO4-  →   Mn2+ +  CO2 
The n factor of Fe2(C2O4)3 would thus be able to be taken as equivalent to the n factor of MnO4- 
Along these lines 
Reactant (n×nf) = Product (n×nf) 
1×6= x×5 
Along these lines x = (6/5) 
Presently on account of ferrous oxalate Fe(C2O4) 
Presently the new response can be composed as: 
MnO4- + FeC2O4 →  Mn2+ +Fe3+ +CO2 
Here additionally Mn is getting diminished from +7 oxidation state to +2 oxidation state. 
Additionally in the compound, ferrous oxalate, Fe is in +2 oxidation state and C is in +3 oxidation state, which are getting changed to +3 and +4 resp. 
So the aggregate n factor for the compound is 1+2 = 3 
Again keeping the counterparts of the reactants and the items to be same on both the sides, 
Y×5=1×3 
Subsequently y = (3/5) 
So the proportion, (x/y) will be (6×5)/(5×3) 
= 2:1 
 
Hope it helps.
Thanks and regards,
Kushagra

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