09276g mn2o3 fm 15788 which is known to be impure is


Question- 0.9276g Mn2O3 (FM 157.88) which is known to be impure is ignited with Na2CO3. This is ioxidized with Bi2O3.

(4Mn)^2 + 6H2O +5Bi2O5 --------> (4MnO4)^- +12H+ +5Bi2O3

(MnO4)^- + (5Fe)^2+ + 8H+ -----------> Mn2+ + (5Fe)^3+ +4H2O

and this has been titrated with13.55 mLof 0.1124M FeSO4

what is the % (w/w) Mn2O3 in the impure sample?

I have no idea how to do this! Thank you for your help!

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Chemistry: 09276g mn2o3 fm 15788 which is known to be impure is
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