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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jcais on March 21, 2006, 06:05:35 PM

Title: Decomposition of Potassium Chlorate
Post by: jcais 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* :-)

Title: Re:Decomposition of Potassium Chlorate
Post by: Borek on March 21, 2006, 06:23:51 PM

Show your work so that someone can show you were you have gone wrong.

Title: Re:Decomposition of Potassium Chlorate
Post by: AWK 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

Title: Re: Decomposition of Potassium Chlorate
Post by: lindeproctor 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)