Can someone explain me how to balance this reaction?
Posted: Thu Jan 05, 2017 7:47 am
(Ru++) +(NH4+) + (Br2) = (Br-) +Ru(NO2)2
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Todo, balance:Di Mezza Alfredo wrote:(Ru++) +(NH4+) + (Br2) = (Br-) +Ru(NO2)2
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Ru{2+} + NH4{+} + Br2 = Br{-} + Ru(NO2)2
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Ru{2+}
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2 NH4{+} = 2 NO2{-}
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Br2 = 2 Br{-}
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2 NH4{+} + 4 H2O = 2 NO2{-}
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Br2 = 2 Br{-}
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2 NH4{+} + 4 H2O = 2 NO2{-} + 16 H{+}
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Br2 = 2 Br{-}
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2 NH4{+} + 16 OH{-} = 2 NO2{-} + 12 H2O
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Br2 = 2 Br{-}
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2 NH4{+} + 16 OH{-} = 2 NO2{-} + 12 H2O + 12 e{-}
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6 Br2 + 12 e{-} = 12 Br{-}
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Ru{2+} + 2 NH4{+} + 16 OH{-} + 6 Br2 = Ru(NO2)2 + 12 Br{-} + 12 H2O
The information available was very usefulGrahamKemp wrote:Todo, balance:Di Mezza Alfredo wrote:(Ru++) +(NH4+) + (Br2) = (Br-) +Ru(NO2)2Remove the spectator ion
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Ru{2+} + NH4{+} + Br2 = Br{-} + Ru(NO2)2
and divide into Red/Ox half cells.Code: Select all
Ru{2+}
Balance O with water
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2 NH4{+} = 2 NO2{-}
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Br2 = 2 Br{-}
Balance H with protons
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2 NH4{+} + 4 H2O = 2 NO2{-}
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Br2 = 2 Br{-}
Migrate to basic
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2 NH4{+} + 4 H2O = 2 NO2{-} + 16 H{+}
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Br2 = 2 Br{-}
Balance charge with electrons
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2 NH4{+} + 16 OH{-} = 2 NO2{-} + 12 H2O
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Br2 = 2 Br{-}
Recombine:
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2 NH4{+} + 16 OH{-} = 2 NO2{-} + 12 H2O + 12 e{-}
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6 Br2 + 12 e{-} = 12 Br{-}
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Ru{2+} + 2 NH4{+} + 16 OH{-} + 6 Br2 = Ru(NO2)2 + 12 Br{-} + 12 H2O