Balance algebraic method

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estambre
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Joined: Sat May 21, 2016 1:17 pm

Balance algebraic method

Post by estambre »

I need help...
Sb2O3+KIO3+HCl+H2O=HSb(OH)6+KCl+ICl
thanks.
GrahamKemp
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Joined: Mon Mar 19, 2012 10:12 pm

Re: Balance algebraic method

Post by GrahamKemp »

estambre wrote:I need help...

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Sb2O3+KIO3+HCl+H2O=HSb(OH)6+KCl+ICl
thanks.
RedOx is happening! Get those spectator ions to safety. They're causing a distraction.

Move the protons and water away too. Strip down to the bare essentials.

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Sb2O3 + IO3{-} + ? = Sb(OH)6{-} + I{+} + ?
(acidic aqueous)

{Note: Yes, I am positive that is iodine cation. That might have been a source of confusion.}

Now we can balance these two half equations by adding

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H{+}, H2O, e
as needed. (Re: acidic aqueous conditions).

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IO3{-} + ?  = I{+} + ? 

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Sb2O3 + ? = 2 Sb(OH)6{-} + ? 
Use water to balance the oxygen.

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IO3{-} + ?  = I{+} + 3 H2O + ?

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Sb2O3 + 9 H2O + ? = 2 Sb(OH)6{-} + ? 
Use protons to balance the hydrogen

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IO3{-} + 6 H{+} + ? = I{+} + 3 H2O + ?

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Sb2O3 + 9 H2O + ? = 2 Sb(OH)6{-} + 6 H{+} +?  
Use electrons to balance the charges.

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IO3{-} + 6 H{+} + 4 e = I{+} + 3 H2O 

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Sb2O3 + 9 H2O = 2 Sb(OH)6{-} + 6 H{+} + 4 e  
Normally we would cross multiply electron counts but they are the same already. So just recombine.

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IO3{-} + Sb2O3 + 6 H2O = I{+} + 2 Sb(OH)6{-}
It is safe for the spectator ions to return.
Now put potassium on the anions, except the antimonite ion which we protonate.
Neutralise remaining positive charges with chlorine.
That is, add

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K{+} + 2 H{+} + 2 Cl{-}
to both sides.

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KIO3 + Sb2O3 + 6 H2O + 2 HCl = ICl + KCl + 2 HSb(OH)6 
Reorder as desired.

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KIO3 + Sb2O3 + 2 HCl + 6 H2O = 2 HSb(OH)6 + KCl + ICl
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