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Topic: Balance: KClO3 + HCl yields KCl + H2O + Cl2 + ClO2  (Read 22422 times)

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capeskim

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Balance: KClO3 + HCl yields KCl + H2O + Cl2 + ClO2
« on: November 03, 2005, 08:11:56 PM »
I have the equation:
KClO3 + HCl yields KCl + H2O + Cl2 + ClO2

I need this to be balanced two different ways.  The coefficient cannot just be a multiple of itsself.

I already have this one:

2KClO3 + 4HCl yields 2KCl + 2H2O + Cl2 + 2ClO2

If someone could please help me out it would be greatly appreciated.
« Last Edit: November 04, 2005, 03:46:08 AM by Mitch »

Offline mike

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Re:Balancing an Equation
« Reply #1 on: November 03, 2005, 08:32:31 PM »
6KClO3 + 28HCl ----> 6KCl + 14H2O + 13Cl2 + 2ClO2
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Offline Borek

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Re:Balancing an Equation
« Reply #2 on: November 03, 2005, 08:50:07 PM »
    2KClO3 +   4HCl ----> 2KCl +   2H2O +    Cl2 + 2ClO2
+  4KClO3 + 24HCl ----> 4KCl + 12H2O + 12Cl2
    ------------------------------------------------------------
    6KClO3 + 28HCl ----> 6KCl + 14H2O + 13Cl2 + 2ClO2
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capeskim

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Re:Balancing an Equation
« Reply #3 on: November 03, 2005, 08:53:43 PM »
Thank you so much. You guys are my heros.  :)
« Last Edit: November 03, 2005, 08:54:23 PM by capeskim »

Offline AWK

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Re:Balancing an Equation
« Reply #4 on: November 04, 2005, 03:27:41 AM »
This is quite simple multisolution redox equation.
You can treat KClO3 as a mixture of K2O and Cl2O5
K2O + Cl2O5 = 2KClO3
Both K2O and Cl2O5 react with HCl
K2O + 2HCl = 2KCl + H2O
Cl2O5 can react with HCl as oxidizer, but depending on (allowed) molar ratio of ClO2 to Cl2 you would to produce, in different proportion Cl2O5/HCl
eg:
Cl2O5 + 10 HCl  = 6Cl2 + 5H2O
Combining two reaction
K2O + 2HCl = 2KCl + H2O
Cl2O5 + 10 HCl  = 6Cl2 + 5H2O
we obtain
2KClO3 + 12HCl = 2KCl + 6Cl2 + 6H2O without ClO2

in this case all coefficients should be divided by a factor 2
KClO3 + 6HCl = KCl + 3Cl2 + 3H2O

other example
3Cl2O5 +  14HCl = 8Cl2 + 4ClO2 + 7H2O

complete reaction
6KClO3 + 20HCl = 6KCl + 8Cl2 + 4ClO2 + 10H2O

and finally
3KClO3 + 10HCl = 3KCl + 4Cl2 + 2ClO2 + 5H2O

and so on

Borek method is equivalent to mine
« Last Edit: November 04, 2005, 03:31:53 AM by AWK »
AWK

Offline Borek

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Re:Balance: KClO3 + HCl yields KCl + H2O + Cl2 + ClO2
« Reply #5 on: November 04, 2005, 04:56:54 AM »
To explain problem further:

This is in fact not one, but two reactions occuring simultaneously (perhaps they are somehow connected due to mechanism):

A: 2KClO3 + 4HCl ----> 2KCl + 2H2O + Cl2 + 2ClO2

B: 4KClO3 + 24HCl ----> 4KCl + 12H2O + 12Cl2

So we have two equations - and we can do mathematical trick. Every linear combination of these two equations will be still valid reaction equation (although it may happen that all coefficients have to be divided by some constant).

So for example we have A+B:

6KClO3 + 28HCl ----> 6KCl + 14H2O + 13Cl2 + 2ClO2

or 2A+3B (divided by 2):

8KClO3 + 40HCl ----> 8KCl + 20H2O + 19Cl2 + 2ClO2

or 7A+5B:

34KClO3 + 148HCl ----> 34KCl + 74H2O + 67Cl2 + 14ClO2

and so on...
« Last Edit: November 04, 2005, 05:01:00 AM by Borek »
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