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Topic: Decomposition of Potassium Chlorate  (Read 7833 times)

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Offline jcais

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Decomposition of Potassium Chlorate
« on: March 21, 2006, 06:05:35 PM »
Hello,

I did the experiment: Decomposition of Potassium Chlorate.

Mass of KClO3 = 1.02g
Mass of unknown product = 0.739g
three possibilties: ???

KClO3 -> KCl + O2

KClO3 -> KClO + O2

2KClO3 -> 2KClO2 + O2

How do I find out the expected masses of each solid product using the mass of KClO3 = 1.02g?

Whatever it is should come close to the unknown product mass.
I set up each problem and got masses of the unknown in the hundreds.

Does anyone know how to set up the dimensional analysis problems? Thank you so much for your *delete me* :-)

Offline Borek

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Re:Decomposition of Potassium Chlorate
« Reply #1 on: March 21, 2006, 06:23:51 PM »
Show your work so that someone can show you were you have gone wrong.
« Last Edit: March 21, 2006, 06:37:41 PM by Borek »
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Offline AWK

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Re:Decomposition of Potassium Chlorate
« Reply #2 on: March 22, 2006, 01:49:01 AM »
Thermal decomposition of KClO3 to KCl goes in two steps:
4KClO3 = 3KClO4 + KCl
Then
KClO4 = KCl + 2O2

Your reaction needs balancing
Quote
KClO3 -> KCl + O2

In your stoichiometric calculation take into account eventual mixture of KCl and KClO4
AWK

Offline lindeproctor

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Re: Decomposition of Potassium Chlorate
« Reply #3 on: October 31, 2007, 10:41:32 PM »
2KCLO3> 2KCL + 302

1.02g KCLO3 * 1mol /122.55g * 2 mol KCL/ 2 mol KCLO3 * 74.551g/mol KCL = grams of product (KCL)

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