Consider the following reaction:
`xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+)3//4" "xMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O`
The values of x, y and z in the reaction are, respectively:

Explanation

The half equations of the reaction are
`MnO_(4)^(-) rarr Mn^(2+)`
`C_(2)O_(4)^(2-) rarr CO_(2)`
The balanced half equations are
`MnO_(4)^(-) + 8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_(2)O`
`C_(2)O_(4)^(2-) rarr 2CO_(2) + 2e^(-)`
On equating number of electrons, we get
`2MnO_(4)^(-) + 16H^(+) + 10e^(-) rarr 2Mn^(2+) + 8H_(2)O`
`5C_(2)O_(4)^(2-) rarr 10CO_(2) + 10e^(-)`
On adding both the equations, we get
`20MnO_(4)^(-) + 16H^(+) + 5C_(2)O_(4)^(2-) rarr 2Mn^(2+) + 16/2H_(2)O + 5 xx 2CO_(2)`
x , y and z are 2, 5 and 16 are respectively.