For the redox reaction
`xMnO_(4)^(ө)+yC_(2)O_(4)^(2-)+zH^(o+)rarrMn^(2+)+CO_(2)+H_(2)O`
the coedfficients `x, y, and `z` are

Options

A. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-), H^(+)),(2,16,5):}`
B. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-), H^(+)),(2,5,16):}`
C. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-), H^(+)),(16,5,2):}`
D. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-), H^(+)),(5,16,2):}`

Answer

B

Explanation

The given redox reaction is
`MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+) to Mn^(2+)+CO_(2)+H_(2)O`
The reaction can be balanced by considering the following steps
Step I Balance the atoms except H and O.
`MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+) to Mn^(2+)+2CO_(2)+H_(2)O`
Step II Write the oxidation number of each atom

Step III Cross multiply by change in oxidation number
`overset(+7)(MnO_(4)^(-) to Mn^(2+), 5e^(-) "gain"`
`overset(+6)(C_(2)O_(4)^(2-) to overset(+8)(2CO_(2)), 2e^(-) "loss"`
`2MnO_(4)^(-)+5C_(2)O_(4)^(-)+H^(+) to 2Mn^(2+)+10CO_(2)+H_(2)O`
Step IV Balance oxygen by adding `H_(2)O` on deficient site.
`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+H^(+) to 2Mn^(2+) +10CO_(2)+8H_(2)O`
Step V Balance hydrogen
`2MnO_(4)^(2-)+5C_(2)O_(4)^(2-)+16H^(+) to 2Mn^(2+)+10CO_(2)+8H_(2)O`
`therefore` The coefficients of the reactants, `MnO_(4)^(-), C_(2)O_(4)^(2-)` ad `H^(+)` are 2, 5 and 16, respectively.