In neutral or faintly alkaline medium,MnO_4^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The balanced chemical equation for the reaction in a neutral or faintly alkaline medium is:$2KMnO_4 + H_2O + KI \rightarrow 2MnO_2 + 2KOH + KIO_3$

Vedclass · Accepted Answer

$1.0$

Vedclass · Answer

(D) The balanced chemical equation for the reaction in a neutral or faintly alkaline medium is:$2KMnO_4 + H_2O + KI ightarrow 2MnO_2 + 2KOH + KIO_3$From the stoichiometry of the balanced equation,$1 \ mole$ of $KI$ reacts with $2 \ moles$ of $KMnO_4$.Given,the amount of $KI = 	ext{Molarity} 	imes 	ext{Volume} = 0.5 \ M 	imes 1 \ L = 0.5 \ moles$.Since $1 \ mole$ of $KI$ requires $2 \ moles$ of $KMnO_4$,then $0.5 \ moles$ of $KI$ will require:$0.5 	imes 2 = 1.0 \ mole$ of $KMnO_4$.

In neutral or faintly alkaline medium,$MnO_4^{-}$ oxidizes $I^{-}$ to iodate. What is the number of moles of $KMnO_4$ required to completely convert $1 \ L$ of $0.5 \ M \ KI$ to iodate?

Similar Questions

$N_2H_4 + IO_3^- + 2H^+ + Cl^- \to ICl + N_2 + 3H_2O$. What is the equivalent weight of the reductant in the above reaction?

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$Pb(NO_3)_2 + H_2SO_4 \longrightarrow PbSO_4 \downarrow + 2HNO_3$

When mercury $(II)$ chloride is treated with excess of $SnCl_2$,the products obtained are

Acidified $KMnO_4$ can be decolourised by

Assign $A, B, C, D$ from the given type of reaction.
$PbO_2 + HNO_3 \text{ (dil.)} \longrightarrow \text{No reaction}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module