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@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.can you make an attempt please?
hint:
moles of electrons involved in oxidation = moles of electrons involved in reduction
sum of both will be the moles of electrons that interact with I2.
i have solved a similar q in the videos btw if you wanna see it (redox reactions)
BUT
n factor = moles of iodine/ moles of electrons involved in EITHER in reduction or oxidation
i did
I used Comproportionation rxn, am i right?
AHHH thats where i made mistake
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found the video
i dont understand hindi sir, but sure will give it a try thanks sir.
awt okok
see if the slides make sense anyway
https://youtu.be/eGACAX7e6_o?list=PL7AAT-ai0VD62sP-5uSYxorGj08lR5FVX&t=1206
this timestamp
you get 5/3
wait wait the n factor of I2 is 5/3
not of KI
KI too right?
why not KI?
Cos its oxidation number changes by 1
Per mole of KI
So n factor is 1
Clo3- n factor is 5
Oh damn ok i get it now.
+solved @iTeachChem
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