The equivalent weight of FeS_2 in the reactio | vedclass

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(B) To find the equivalent weight,we determine the change in oxidation state of $FeS_2$ in the reaction $FeS_2 	o Fe_2O_3 + SO_2$.In $FeS_2$,the oxid

Vedclass · Accepted Answer

$M/11$

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(B) To find the equivalent weight,we determine the change in oxidation state of $FeS_2$ in the reaction $FeS_2 	o Fe_2O_3 + SO_2$.In $FeS_2$,the oxidation state of $Fe$ is $+2$ and $S$ is $-1$ (as $S_2^{2-}$).In $Fe_2O_3$,the oxidation state of $Fe$ is $+3$.In $SO_2$,the oxidation state of $S$ is $+4$.Change in oxidation state of $Fe$: $1 	imes (+3 - (+2)) = +1$.Change in oxidation state of $S$: $2 	imes (+4 - (-1)) = 2 	imes 5 = +10$.Total change in oxidation state (n-factor) = $1 + 10 = 11$.Equivalent weight = $\frac{	ext{Molecular weight}}{	ext{n-factor}} = \frac{M}{11}$.

List-$I$ (Substance)	List-$II$ (Equivalent weight)
$A. Na_2CO_3$	$I. M/5$
$B. KMnO_4 / H^+$	$II. M/3$
$C. K_2Cr_2O_7 / H^+$	$III. M/2$
$D. KMnO_4 / H_2O$	$IV. M/6$

The equivalent weight of $FeS_2$ in the reaction $FeS_2 \to Fe_2O_3 + SO_2$ is (where $M$ is the molecular weight of $FeS_2$):

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Match the following substances in List-$I$ with their equivalent weights in List-$II$ ($M$ = Formula weight):

List-$I$ (Substance) List-$II$ (Equivalent weight)

$A. Na_2CO_3$ $I. M/5$

$B. KMnO_4 / H^+$ $II. M/3$

$C. K_2Cr_2O_7 / H^+$ $III. M/2$

$D. KMnO_4 / H_2O$ $IV. M/6$

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