How many grams of potassium dichromate are required to oxidise 20.0 g of `Fe^(2+)` in `FeSO_4 " to " Fe^(3+)` if the reaction is carried out in an acidic solution ?

Explanation

When dissolved in water , `K_2Cr_2O_7` dissociates into `K^+ and Cr_2O_7^(2-)` ions and `FeSO_4` dissociates into `Fe^(2+) and SO_4^(2+)` ions. The skeleton equation is :
`Fe^(2+) + Cr_2O_7^(2-) rarr Fe^(3+) +2Cr^(3+)`
The balanced equation is
`Cr_2O_7^(2-)+6Fe^(2+)+14H^(+)rarr6Fe^(3+)+2Cr^(3+)+7H_2O`
Formula weight of
`K_2Cr_2O_7=2xx39+2xx52+7xx16=294`
Formula weight of `FeSO_4`
`=56+32+4xx16=152`
`:.` 1 mol `K_2Cr_2O_7=6 mol FeSO_4`
`294 " "6xx152`
`6xx152` g of `FeSO_4` require `K_2Cr_2O_7 = 294 g`
20 g of `FeSO_4` require `K_2Cr_2O_7`
`=(294)xx(6xx152)xx20=6.45g`
`K_2Cr_2O_7` required to oxidise 20 g of `FeSO_4`
`=6.45 g `