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Q.

The number of moles of Question Image and Question Image separately required to oxidise one mole of FeC2O4 each in acidic medium respectively

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a

0.5 ; 0.6

b

0.6; 0.4

c

0.4 ; 0.5

d

0.6 ; 0.5

answer is D.

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Detailed Solution

n-factor for MnO4- = 5 (MnO4- → Mn+2)
n-factor for Cr2O7-2 = 6 (Cr2O7-2 → Cr+3)
n-factor for FeC2O4 = 3 (Fe+2 →Fe+3 & C2O4-2 →CO2)
 

large frac{{MnO_4^ - }}{5}frac{{Fe{C_2}{O_4}}}{3}


n-factor
Criss cross
 

 

3moles of  MnO4- oxidizes 5 mole of FeC2O4
'X' mole of MnO4- oxidizes 1mole of FeC2O4
 

large X_1;=;frac{3}{5};=;0.6


 

large frac{{C{r_2}O_7^{ - 2}}}{6};frac{{Fe{C_2}{O_4}}}{3}


n-factor
Criss - cross method
3 moles of Cr2O7-2 oxidizes 6 mole of FeC2O4
'X2' mole of Cr2O7-2 oxidizes 1 mole of FeC2O4
 

large X_2;=;frac{3}{6};=;0.5
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