`KMnO_4` reacts with oxalic acid according to the equation `2MnO_4^(-) + 5C_2 O_4^(2-) + 16H^+ rarr 2Mn^(2+) + 10CO_2 + 8H_2O` Here, `20mL` of `0.1M KMnO_4` is equivalent to :

KMnO_(4) reacts with oxalic acid according to the equation 2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) to 2Mn^(2+)+10Co_(2)+8H_(2)O Here, 20mL of 1.0M KMnO_(4) is equivalent to:

Complete the reaction: 2MnO_(4)^(-)+5C_(2)O_(4)^(2-) +16H^(+) to

In the balanced equation MnO_(4)^(-)+H^(+)+C_(2)O_(4)^(2-)rarr Mn^(2+)+CO_(2)+H_(2)O , the moles of CO_(2) formed are :-

Oxalic acid, H_(2)C_(2)O_(4) , reacts with paramagnet ion according to the balanced equation 5H_(2)C_(2)O_(4)(aq)+2MnO_(4)^(-)(aq)+6H^(+)(aq)hArr2Mn^(2+)(aq)+10CO_(2)(g)+8H_(2)O(l) . The volume in mL of 0.0162 M KMnO_(4) solution required to react with 25.0 mL of 0.022 M H_(2)C_(2)O_(4) solution is

The value of n in the following half equation is MnO4^(-) + 2H_2O + ne rarr MnO_2 + 4O'H

In the following reaction 2MnO_(4^(-))+5H_(2)O_(2)^(18)+6H^(+)rarr2Mn^(2+) + 8H_(2)O+5O_(2) The radioactive oxygen will appear in :

For the redox reaction, MnO_4^- + C_2 O_4^(2-) + H^+ rarr Mn^(2+) + CO_2 + H_2 O the correct coefficients of the reactants for the balanced reaction are respectively MnO_4^-, C_2 O_4^-, H^+ :

KMnO_(4) (m.w.=158) oxidises oxalic acid in acid medium to CO_(2) and water as follows : 5C_(2)O_(4)^(2-)+2MnO_(4)^(-)+16H^(+) to 10 CO_(2)+2Mn^(2+)+8H_(2)O What is the equivalent weigth of KMnO_(4) ?

In the reaction KMnO_(4) + H_(2)SO_(4) + H_(2)C_(2)O_(4) to products, Mn^(2+) ions act as:

K_(2)MnO_(4)+H^(+) to KMnO_(4)+MnO_(2)darr