1 mole of equimolar mixture of Fe_ 2 (C_ 2 O _ 4 )_ 3 and FeC_ 2 O _ 4 requires X moles of KMnO_ 4 in acid medium for complete reaciton. The value of X is:

Correct answer: A. 0.9 cases Fe _ 2 ( C _ 2 O _ 4 ) _ 3 & ⟶ & 2 Fe ^ 3 + + 6 CO _ 2 + 6 - e & (n Factor = 6) Fe ( C _ 2 O _ 4 ) & ⟶ & Fe ^ 3 + + 2 CO _ 2 + 3 - e & (n Factor = 3) cases Meqs of [ Fe_ 2 (C_ 2 O _ 4 )_ 3 + Fe(C_ 2 O _ 4 )]=0.5×6+0.5×3 =4.5Eqs. =4.5Eqs. of KMnO_ 4 = 4.5 5 =0.9 moles of KMnO_ 4 .

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Redox Reactions

NTA Abhyas 2020 · Chemistry · Redox Reactions

1 mole of equimolar mixture of Fe_ 2 (C_ 2 O _ 4 )_ 3 and FeC_ 2 O...

\text{1}

mole of equimolar mixture of

{\text{Fe}_{2} {{\text{(C}_{2} {O}_{4} \text{)}_{3}

and

{\text{FeC}_{2} {O}_{4}

requires

\text{X}

moles of

{\text{KMnO}_{4}

in acid medium for complete reaciton. The value of

\text{X}

is:

Options

A0.9
B0.6
C1.2
D0.8

Correct answer

A. 0.9

Step-by-step solution

\begin{cases}{Fe}_{2} {\left({C}_{2} {O}_{4}\right)}_{3} & ⟶ & 2 {Fe}^{3 +} + 6 {CO}_{2} + 6 \overset{-}{e} & \left(\text{n Factor = 6}\right) \\ \text{Fe} \left({C}_{2} {O}_{4}\right) & ⟶ & {Fe}^{3 +} + 2 {CO}_{2} + 3 \overset{-}{e} & \left(\text{n Factor = 3}\right)\end{cases}

<br />Meqs of \left[{\text{Fe}_{2} {{\text{(C}_{2} {O}_{4} \text{)}_{3} \text{+}{\text{Fe(C}_{2} {O}_{4} \text{)}\right]\text{=}\text{0}\text{.5}\text{\times}\text{6}\text{+}\text{0}\text{.5}\text{\times}\text{3}

\text{=}\text{4}\text{.5}\text{Eqs}\text{.}

\text{=}\text{4}\text{.5}\text{Eqs}\text{.}

{\text{KMnO}_{4}

\text{=}\dfrac{\text{4}\text{.5}}{5} \text{=}\text{0}\text{.9}

moles of

{\text{KMnO}_{4} \text{.}

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