The number of moles of I 2 produced when when 500mL,0.5M KMnO 4 reacts with sufficient KI in acidic medium is

The reaction involves MnO₄⁻ oxidizing I⁻ to I₂. The balanced redox reaction is: 2 M n O 4 − + 10 I − + 16 H + → 2 M n 2 + + 5 I 2 + 8 H 2 O Moles of KMnO₄ = 0.5 L × 0.5 m o l / L = 0.25 m o l From stoichiometry, 2 moles of MnO₄⁻ produce 5 moles of I₂. Moles of I₂ = ( 0.25 m o lM n O 4 − ) × 2 m o lM n O 4 − 5 m o l I 2 = 0.625 m o l

Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsProblem involving Equivalent ConceptMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

KMnO₄ act as oxidising agent in acidic, basic and neutral medium.

In acidic medium KMnO₄ oxidises I^- to I₂ , into C₂O₄^2- into CO₂

Answer the following questions.

The number of moles of I₂ produced when when 500mL,0.5M KMnO₄ reacts with sufficient KI in acidic medium is

Answer:

0.62

Solution:

The reaction involves MnO₄⁻ oxidizing I⁻ to I₂.

The balanced redox reaction is:

$2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 M n^{2 +} + 5 I_{2} + 8 H_{2} O$

Moles of KMnO₄ = $0.5 L \times 0.5 m o l / L = 0.25 m o l$

From stoichiometry, 2 moles of MnO₄⁻ produce 5 moles of I₂.

Moles of I₂ = $(0.25 m o lM n O_{4}^{-}) \times \frac{5 m o l I _{2}}{2 m o lM n O _{4}^{-}} = 0.625 m o l$

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Problem involving Equivalent Concept

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...