How many grams of `K_(2)Cr_(2)O_(7)` are required to oxidize `Fe^(2+)` present in 15.2 g of `FeSO_(4) to Fe^(3+)` if the reaction is carried out in an acidic medium ?

Answer

4.9 g

Explanation

The balanced chemical equation for the redox reaction is
`K_(2)Cr_(2)O_(7)+6FeSO_(4)+7H_(2)SO_(45)rarrK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+7H_(2)`
From the balanced equation it is clear that
6 Moles of `FeSO_(4)=1 "Mole of" K_(2)Cr_(2)O_(7) or 6xx152 g "for" FeSO_(4)` are oxidised by `K_(2)Cr_(2)O_(7)=294` g
152 g of `FeSO_(4)` are oxidised by `K_(2)Cr_(2)O_(7)=(294)/(6xx152)xx15.2=4.9 g`