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Balance the following equation by oxidation number method Al+KMnO_(4)+H_(2)SO_(4) to Al_(2)(SO_(4))_(3)+K_(2)SO_(4)+MnSO_(4)+H_(2)O |
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Answer» Solution :Writing oxidation numbers of all atoms, `overset(0)(Al)+overset(+1 +7 -2)(KMnO_(4))+overset(+1 +6 -2)(H_(2)SO_(4)) to overset(+3 +6 -2)(Al_(2)(SO_(4))_(3))+overset(+1 +6 -2)(K_(2)SO_(4))+overset(+2 +6 -2)(MnSO_(4))0+overset(+1 -2)(H_(2)O)` The oxidation numbers of Al and Mn have changed `overset(0)(Al) to overset(+3)(Al_(2))(SO_(4))_(3)....(i)` `overset(+7)(KMnO_(4)) to overset(+2)(MnSO_(4))....(II)` Increase in Ox. no. of Mn=5 units per `KMnO_(4)` molecule Multiply eq. (i) by 10 and eq. (ii) by 6 as to make increase and decrease equal. `10Al+6KMnO_(4) to 5Al_(2)(SO_(4))_(3)+6MnSO_(4)+3K_(2)SO_(4)` To balance `SO_(4)^(2-)` ions, `24H_(2)SO_(4)` molecules be added on LHS. `10Al+6KMnO_(4)+24H_(2)SO_(4) to 5Al_(2)(SO_(4))_(3)+6MnSO_(4)+3K_(2)SO_(4)` To balance hdyrogen and oxygen, 24 `H_(2)O` molecules be added on RHS. Hence, the balanced equation is `10Al+6KMnO_(4)+24H_(2)SO_(4) to 5Al_(2)(SO_(4))_(3)+6MnSO_(4)+3K_(2)SO_(4)+2H_(2)O` |
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