If the molecular weights of Na_2S_2O_3 and I_ | vedclass

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

Question

(B) The given reaction is: $2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$For $S_2O_3^{2-}$: The oxidation state of $S$ changes from $+2$ to $

Vedclass · Accepted Answer

$M_1, M_2 / 2$

Vedclass · Answer

(B) The given reaction is: $2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$For $S_2O_3^{2-}$: The oxidation state of $S$ changes from $+2$ to $+2.5$. The change in oxidation number per $S$ atom is $0.5$. Since there are $2$ sulfur atoms in $S_2O_3^{2-}$, the total change per mole is $0.5 	imes 2 = 1$. Thus, the $n$-factor is $1$. Equivalent weight = $\frac{M_1}{1} = M_1$.For $I_2$: The oxidation state of $I$ changes from $0$ to $-1$. The change in oxidation number per $I$ atom is $1$. Since there are $2$ iodine atoms in $I_2$, the total change per mole is $1 	imes 2 = 2$. Thus, the $n$-factor is $2$. Equivalent weight = $\frac{M_2}{2}$.

If the molecular weights of $Na_2S_2O_3$ and $I_2$ are $M_1$ and $M_2$ respectively, then what will be the equivalent weights of $Na_2S_2O_3$ and $I_2$ in the following reaction?
$2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$

Similar Questions

The number of moles of $KMnO_4$ reduced by one mole of $KI$ in alkaline medium is

The equivalent weight of $KIO_3$ in the reaction $2Cr(OH)_3 + 4OH^{-} + KIO_3 \to 2CrO_4^{2-} + 5H_2O + KI$ is

Assign $A, B, C, D$ from the given type of reaction.
$Sn + PCl_5 \xrightarrow{\Delta} SnCl_4 + PCl_3$

In the equation,$BiO_3^{-} + 6H^{+} + xe^{-} \rightarrow Bi^{3+} + 3H_2O$. What is the value of $x$?

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$2NaI + MnO_2 + 2H_2SO_4 \to MnSO_4 + Na_2SO_4 + I_2 + 2H_2O$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If the molecular weights of $Na_2S_2O_3$ and $I_2$ are $M_1$ and $M_2$ respectively, then what will be the equivalent weights of $Na_2S_2O_3$ and $I_2$ in the following reaction?$2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$