Exam

JEE Main

Year

2019

Date

8 April

Shift

Morning Shift

Subject

Chemistry

Chapter

Redox Reactions

JEE Main 2019 · 8 April · Morning Shift · Chemistry · Redox Reactions

In order to oxidize a mixture of one mole of each of Fe C _ 2 O _ 4...

In order to oxidize a mixture of one mole of each of $Fe{C}_{2}{O}_{4}, F{e}_{2}{( {C}_{2}{O}_{4} )}_{3}, FeS{O}_{4}$ and $F{e}_{2}{( S{O}_{4} )}_{3}$ in acidic medium, the number of moles of $KMn{O}_{4}$ is:

Step-by-step solution

$Eq . KMn{O}_{4} ⇒ Eq . FeS{O}_{4} + Eq . F{e}_{2}{\left(S{O}_{4}\right)}_{3} + EqFe{C}_{2}{O}_{4} + Eq . F{e}_{2}{\left({C}_{2}{O}_{4}\right)}_{3}$
$KMn{O}_{4} ⇒ Mn{O}_{4}^{-}$
$5{e}^{-} + 8{H}^{+} + Mn{O}_{4}^{-} ⇒ M{n}^{+ 2} + 4{H}_{2}O$ $\text{V.F} = 5$

So $Eq . KMn{O}_{4} ⇒ Eq . FeS{O}_{4} + Eq . Fe{C}_{2}{O}_{4} + Eq . F{e}_{2}{\left({C}_{2}{O}_{4}\right)}_{3}$
$n \times 5 = 1 \times 1 + 1 \times 0 + 1 \times 3 + 1 \times 6$
$n \times 5 = 1 + 3 + 6$
$n = \dfrac{10}{5} = 2mole$