The values of `X, Y` and `Z`s in the reaction are repectively:
`XMnO_(4)^(-)+YH_(2)SO_(4)rarr2Mn^(2+)+5H_(2)O+9O_(2)+Ze`

Options

A. 2,6,6
B. 5,2,9
C. 3,5,5
D. 2,6,6

Answer

A

Explanation

For balancing Mn on both sides, multiply `MnO_(4)^(-)` by2. To balance `5H_(2)O`, multiply `H_(2)O_(2)`by 5
`2MnO_(4)^(-)+5H_(2)O_(2)rarr2Mn^(2+)+5H_(2)O+9O_(2)+6e^(-)`
To balance charge on both sides `6e^(-)` have been added on right hand side.