2g Feso4 is completely oxidised by 0.05M kmno4.what volume of kmno4 is required?

Asked by nareshrajpurohit43109 | 22nd May, 2020, 11:18: AM

Expert Answer:

The reaction is,
 
   10FeSO4   +   2KMnO +  8H2SO4 → KSO4 + 2MnSO4 + 5Fe2(SO4) + 8H2O
(10×151.8)       (2×158)       
 
 
10×151.8 gm of FeSO4 require KMnO4 = 2×158 g
 
2 gm of FeSO4 will require KMnO4 = fraction numerator 2 cross times 2 cross times 158 over denominator 10 cross times 151.8 end fraction
 
Let V ml of 0.05 M KMnO4 solution is required,
 
Amount of KMnO4 in the solution  fraction numerator 158 cross times 0.05 over denominator 1000 end fraction cross times straight V
Thus, space fraction numerator 158 cross times 0.05 cross times straight V over denominator 1000 end fraction

equals space fraction numerator 2 cross times 158 cross times 2 over denominator 10 cross times 151.8 end fraction

straight V equals space 52.7 space ml

Answered by Varsha | 22nd May, 2020, 04:40: PM