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### Topic: Balancing equations  (Read 8981 times)

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#### samanthaln86

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##### Balancing equations
« on: January 12, 2005, 08:37:55 PM »
I cannot for the life of me balance this equation.

Ce(IO3)4  + H2C2O4    –>    Ce(C2O4)3  +  I2  + H2O   +   CO2
« Last Edit: January 12, 2005, 08:43:10 PM by samanthaln86 »

#### samanthaln86

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##### Re:Balancing equations
« Reply #1 on: January 12, 2005, 08:41:05 PM »
and yes i have tried to do it over and over, maybe im just looking over something

#### Tetrahedrite

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##### Re:Balancing equations
« Reply #2 on: January 12, 2005, 09:53:45 PM »
It is a redox reaction, you have to balance each half separately or it will be too hard.
Hint I(oxidation state 5+) >>>>>>>I (oxidation state 0)
and
Ce4+ >>>>>>>>>Ce3+

#### Mitch

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##### Re:Balancing equations
« Reply #3 on: January 12, 2005, 10:45:31 PM »
I was bored so you get lucky today. Start with cerium and it falls out eventually, the trick is to do iodine next and then "(O3)" then hydrogen and finally the CO2s

2 Ce(IO3)4  + 24H2C2O4    –>    2Ce(C2O4)3  +  4I2  + 24H2O  +  36CO2
« Last Edit: January 13, 2005, 02:07:08 AM by Mitch »
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
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#### AWK

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##### Re:Balancing equations
« Reply #4 on: January 13, 2005, 01:59:12 AM »
Cerium is the oxidant or not in this reaction (probably is), but cerium oxalate should be Ce2(C2O4)3 in the former case or Ce(C2O4)2 in the later one.
These reaction can be balanced even without procedure for redox reaction taking into account hydrogen atoms on the left side and water on the right side.
Follow up my way of thinking!
You need 2 cerium atoms on the right side. Put 2 on the left one.
2Ce(IO3)4  + H2C2O4    –>    Ce2(C2O4)3  +  I2  + H2O  +  CO2
you have 2 cerium atoms, 8 iodine atoms and 24 oxygen atoms that need 48 hydrogen atoms to form water, hence you need 24 molecules of oxalic acid, 3 of them form cerium(III) oxalate, the rest molecules of oxalic acid forms carbon dioxide. So the result is
2Ce(IO3)4  + 24H2C2O4    –>    Ce2(C2O4)3  +  4I2  + 24H2O  +  42CO2

In the second case, thinking analogously:
Ce(IO3)4  + 12H2C2O4    –>    Ce(C2O4)2  +  2I2  + 12H2O  +  20CO2
« Last Edit: January 13, 2005, 02:03:53 AM by AWK »
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#### jdurg

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##### Re:Balancing equations
« Reply #5 on: January 13, 2005, 08:44:39 AM »
Maybe it's because I was/am such a geek, but I always enjoyed balancing chemical reactions.  I thought it was pretty neat.
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#### samanthaln86

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##### Re:Balancing equations
« Reply #6 on: January 13, 2005, 08:53:42 AM »
thanks so much for everyones *delete me*!

#### AWK

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##### Re:Balancing equations
« Reply #7 on: January 13, 2005, 09:02:15 AM »
Specially for Jdurg, an easy example, just for enjoying.
Cr2P2O7 + I2 = Cr2O3 + CrP3O9 + PI3
« Last Edit: January 13, 2005, 09:04:01 AM by AWK »
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#### jdurg

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##### Re:Balancing equations
« Reply #8 on: January 13, 2005, 09:57:38 AM »
This one wasn't too bad.  It was just a matter of figuring out the oxidation states that took some effort.  But in the end, the resulting balanced equation is:

15Cr2P2O7 + 9I2 -> 11Cr2O3 + 8CrP3O9 +6PI3.
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