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(vii) H_(2)C_(2)O_(4) +KMnO_(4)+H_(2)SO_(4)to H_(2)SO_(4)+MnSO_(3)+CO_(2)+H_(2)O. |
Answer» Solution : Equalise the INCREASE/ decrease in O N by multiplaying Cu species by 5 and Mn species by 1 `5H_(2)C_(2)O_(4)+KMnO_(4)+H_(2)SO_(4)to K_(2)SO_(4)MnSO_(4)+5CO_(2)+H_(2)O` Balance all other atoms except H and O atoms `5H_(2)C_(2)O_(4)+2KMnO_(4)+3H_(2)SO_(4)to K_(2)SO_(4) +2MnSO_(4)+10CO_(2)+H_(2)O` Balance O ATOM by adding `H_2O` on the side falling SHORT of oxygen atoms. `5H_(2)C_(2)O_(4)+2KMnO_(4)+3H_(2)SO_(4)to K_(2)SO_(4)+2MnSO_(4)+10CO_(2)+H_(2)O+7H_(2)O` The balanced equation in `5H_(2)C_2O_(4)+2KMnO_(4)+3H_(2)SO_(4)to K_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O`. |
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