How do you balance $K C l O_{3} (s) \to K C l (s) + O_{2} (g)$ $KClO_3(s) -> KCl(s) + O_2(g)$ ?

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How do you balance $K C l O_{3} (s) \to K C l (s) + O_{2} (g)$ $KClO_3(s) -> KCl(s) + O_2(g)$ ?

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Answer 1 · 2018-04-22T04:02:32

Using oxidation states.

Explanation:

We have to realize that this is a redox reaction.
Chlorine is being reduced while oxygen is being oxidized.
Chlorine oxidation state is $+ 5$ $+5$ in $K C l O_{3}$ $KClO_3$ and is reduced to $C l -$ $Cl-$ which has an oxidation state of $- 1$ $-1$ .
Oxygen oxidation state is $- 2$ $-2$ in $K C l O_{3}$ $KClO_3$ and is oxidised to $O_{2}$ $O_2$ which has an oxidation state of $0$ $0$ .
Write Half Reactions.
$C l^{5 +} + 6 e = C l^{-}$ $Cl^(5+) + 6e = Cl^-$
$2 O^{2 -} = O_{2} + 4 e$ $2O^(2-) = O_2 + 4e$
Balance electrons in accordance to conservation of charges.
$4 C l^{5 +} + 24 e = 4 C l^{-}$ $4Cl^(5+) + 24e = 4Cl^-$
$12 O^{2 -} = 6 O_{2} + 24 e$ $12O^(2-) = 6O_2 + 24e$

Thus interpreting this means that there is $4 K C l O_{3}$ $4KClO_3$ , $4 K C l$ $4KCl$ , $6 O_{2}$ $6O_2$ . This balances both atoms and charges making it fully balanced.
Thus simplify the equation by a factor of 2,
$2 K C l O_{3} = 2 K C l + 3 O 2$ $2KClO_3 = 2KCl + 3O2$

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How do you balance $K C l O_{3} (s) \to K C l (s) + O_{2} (g)$ $KClO_3(s) -> KCl(s) + O_2(g)$ ?

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How do you balance KClO_3(s) -> KCl(s) + O_2(g)KClO3(s)→KCl(s)+O2(g)KClO_3(s) -> KCl(s) + O_2(g)?

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