A) \[2/5\]
B) \[3/5\]
C) \[4/5\]
D) \[1\]
Correct Answer: B
Solution :
| \[{{\operatorname{KMnO}}_{4}}+{{\operatorname{FeC}}_{2}}{{\operatorname{O}}_{4}}\xrightarrow[{}]{}{{\operatorname{Fe}}^{3+}}\]\[+2C{{O}_{2}}+M{{n}^{2+}}\] |
| So half reaction, \[\operatorname{MnO}_{4}^{-}\xrightarrow[{}]{}{{\operatorname{Mn}}^{2+}}\left( \operatorname{decrease}\,in\,O.N.=5 \right)\]\[{{\operatorname{FeC}}_{2}}{{O}_{4}}\xrightarrow[{}]{}F{{e}^{3+}}+2C{{O}_{2}}\] \[\left( increase\,in\,O.N=1 \right)\] |
| Now equating the change in O.N. and then by adding both half reaction we get |
| \[5{{\operatorname{FeC}}_{2}}{{O}_{4}}+MnO_{4}^{-}\xrightarrow[{}]{}\]\[5F{{e}^{3+}}+10C{{O}_{2}}+M{{n}^{2+}}\] |
| On balancing equation,\[3MnO_{4}^{1}+5Fe{{C}_{2}}{{O}_{4}}+24{{H}^{+}}\xrightarrow[{}]{}\]\[5F{{e}^{3+}}+3M{{n}^{2+}}+10C{{O}_{2}}+12{{H}_{2}}O\] \[\therefore 3\operatorname{moles}\,of\,KMn{{O}_{4}}=5\operatorname{moles}\,of\,{{\operatorname{FeC}}_{2}}{{O}_{4}}\]\[1mole\,Fe{{C}_{2}}{{\operatorname{O}}_{4}}=\frac{3}{5}moles\,\operatorname{of}\,KMn{{O}_{4}}\] |
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