`H_(3)PO_(2)+CuSO_(4) to Cu darr+H_(3)PO_(4)+HNO_(3)`

Options

A. The number of moles of `Cr_(2)O_(7)^(2-)` required to oxidise 6 mol of `Cu_(3)P` to `CuSO_(4)` and `H_(3)PO_(4)` is 11 mol.
B. The number of moles of `H_(2)SO_(4)` used in the reaction is 62
C. The number of moles of `Cr_(2)(SO_(4))_(3)` formed in the reaction is 11
D. The number of moles of `K_(2)SO_(4)` formed in the reaction is 11

Answer

A, B, C, D

Explanation

`6Cu_(3)P +11K_(2)Cr_(2)O_(7) + 62H_(2)SO_(4) rarr 18CuSO_(4)`
`+6H_(3)PO_(4)+53 H_(2)O +11Cr_(2) (SO_(4))_(3) +11 K_(2)SO_(4)`
Number of moles of `H_(2)SO_(4) =`used =
`62 H_(2)SO_(4) (=124 H^(+) = 62 SO_(4)^(2-))`
Number of moles of `CuSO_(4) =` formed =
`18CuSO_(4) = (=18 SCO_(4)^(2-))`
Number of moles of `Cr_(2)(SO_(4))_(3)` formed `=11 Cr_(2)(SO_(4))_(3) = (33SO_(4)^(2-))`
Total number of `SO_(4)^(2-)` ion in reactant `=62`
Total number of `SO_(4)^(2-)` ion in product `=18 +33 =51`
Rest 11 mol of `SO_(4)^(2-)` ion in the reactant react with `11//2` mol of `K^(+)` ion to give 11 mol of `K_(2)SO_(4)`. Net redox reaction is: