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In this reaction,
$$\ce{KIO_3 + KI -> I_2}$$ According to law of chemical equivalence, shouldn't $m_{eq}$ of $\ce{KIO_3}$ = $m_{eq}$ of $\ce{KI}$ = $m_{eq}$ of $\ce{I_2}$?
However, the book that I was reading from, said:
Total $m_{eq}$. of $\ce{I_2}$ = $m_{eq}$ of $\ce{I_2}$ (from $\ce{KIO_3}$) + $m_{eq}$ of $\ce{I_2}$ (from $\ce{KI}$).
Why is it so?

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This Reaction is a Disproportionation Reaction, that is, a reaction involving both oxidation and reduction of a single species, in this case : Iodine. This is not one reaction but a combination of two reactions: $$\ce{KIO3 -> I2}$$ And $$\ce{KI -> I2}$$ Therefore, by applying law of chemical equivalence in both reactions and adding the equations :

$m_{eq}$ of $\ce{I_2}$ (from both reactions) = $m_{eq}$ of $\ce{I_2}$ (from $\ce{KIO3}$) + $m_{eq}$ of $\ce{I_2}$ (from $\ce{KI}$)

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  • $\begingroup$ Welcome to chem.SE! You might also write the reactions completely to avoid future confusions from the readers of this answer. $\endgroup$ – M.A.R. Jun 7 '15 at 19:18
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    $\begingroup$ Technically it is a comproportionation (two different oxidation states going to the same oxidation state), not a disproportionation (the same oxidation state going to different oxidation states, such as $\ce{Cl2 -> Cl- + OCl-}$. I think it's a trivial matter though. $\endgroup$ – orthocresol Jun 10 '15 at 20:18

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