Balance algebraic method
Posted: Sat May 21, 2016 1:24 pm
I need help...
Sb2O3+KIO3+HCl+H2O=HSb(OH)6+KCl+ICl
thanks.
Sb2O3+KIO3+HCl+H2O=HSb(OH)6+KCl+ICl
thanks.
RedOx is happening! Get those spectator ions to safety. They're causing a distraction.estambre wrote:I need help...thanks.Code: Select all
Sb2O3+KIO3+HCl+H2O=HSb(OH)6+KCl+ICl
Move the protons and water away too. Strip down to the bare essentials.
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Sb2O3 + IO3{-} + ? = Sb(OH)6{-} + I{+} + ?{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, eCode: Select all
IO3{-} + ? = I{+} + ? Code: Select all
Sb2O3 + ? = 2 Sb(OH)6{-} + ? Code: Select all
IO3{-} + ? = I{+} + 3 H2O + ?Code: Select all
Sb2O3 + 9 H2O + ? = 2 Sb(OH)6{-} + ? Code: Select all
IO3{-} + 6 H{+} + ? = I{+} + 3 H2O + ?Code: Select all
Sb2O3 + 9 H2O + ? = 2 Sb(OH)6{-} + 6 H{+} +? Code: Select all
IO3{-} + 6 H{+} + 4 e = I{+} + 3 H2O Code: Select all
Sb2O3 + 9 H2O = 2 Sb(OH)6{-} + 6 H{+} + 4 e Code: Select all
IO3{-} + Sb2O3 + 6 H2O = I{+} + 2 Sb(OH)6{-}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{-}Code: Select all
KIO3 + Sb2O3 + 6 H2O + 2 HCl = ICl + KCl + 2 HSb(OH)6 Code: Select all
KIO3 + Sb2O3 + 2 HCl + 6 H2O = 2 HSb(OH)6 + KCl + ICl