How many grams of potassium dichromate are required to oxidise 15.2 g of `FeSO_(4)` in acidic medium?

Explanation

To write balanced chemical equation of the redox reaction
`Cr_(2)O_(7)^(2-)+14 H^(+)+6 e^(-) rarr 2Cr^(3+)+7 H_(2)O`
`Fe^(2+) rarr Fe^(3+)+e^(-)]xx6`
`Cr_(2)O_(7)^(2-)+6 Fe^(2+)+014 H^(+) rarr 2Cr^(3+)+6Fe^(3+)+7 H_(2)O`
step 2 To calculate the amount of `K_(2)Cr_(2)O_(7)` required
It is evident from balanced equation that
1 mole of `Cr_(2)O_(7)^(2-)=6 "moles of" Fe^(2+) "or 1 mole" k_(2)Cr_(2)O_(7)=6 "moles" FeSO_(4)`
But 1 mole of `K_(2)Cr_(2)O_(7)=2xx39+2xx52+16xx7=294g`
and 6 moles of `FeSO_(4)=6xx(56+32+16+45)=6xx152g FeSO_(4)`
Now `6xx152 g "if" FeSO_(4)` are oxidised by `K_(2)Cr_(2)O_(7)=(294xx152)/(6xx152)=4.9g`