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Topic: Problem of the week - 15/10/2012  (Read 8659 times)

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Borek

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Problem of the week - 15/10/2012
« on: October 15, 2012, 06:23:29 AM »
Following five reactions have one common reactant and one common product:

A + HCl F + CrCl3 + H + I
B + HCl F + G + H + I
C + HCl F + H + I
D + HCl G + H + I
E + HCl H + I

Identify all substances, knowing that C contains 31.83% of metal by mass.
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Rutherford

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Re: Problem of the week - 15/10/2012
« Reply #1 on: October 16, 2012, 12:51:41 PM »
Okay, I think it's time to write my attempt:
A=K2Cr2O7
B=KMnO4
C=KClO3
D=MnO2
E=HClO
F=KCl
G=MnCl2
H=H2O
I=Cl2

Borek

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Re: Problem of the week - 15/10/2012
« Reply #2 on: October 18, 2012, 06:27:16 AM »
There is an alternative answer to E.
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SchrÃ¶dinger

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Re: Problem of the week - 15/10/2012
« Reply #3 on: October 18, 2012, 06:47:13 AM »
H2O2? Oxidizing HCl to Cl2 and hence itself getting reduced to H2O?
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Borek

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Re: Problem of the week - 15/10/2012
« Reply #4 on: October 18, 2012, 06:54:50 AM »
Even simpler than that. Reaction I am thinking about was part of an industrial process.
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Rutherford

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Re: Problem of the week - 15/10/2012
« Reply #5 on: October 18, 2012, 08:11:26 AM »
E=O2?

Borek

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Re: Problem of the week - 15/10/2012
« Reply #6 on: October 18, 2012, 08:15:47 AM »
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XGen

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Re: Problem of the week - 15/10/2012
« Reply #7 on: October 19, 2012, 04:55:16 PM »
If it's not an inconvenience, could you explain your beginning approach when tackling this problem? I'm interested to hear about it

Borek

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Re: Problem of the week - 15/10/2012
« Reply #8 on: October 20, 2012, 01:53:09 PM »
Seeing how in all cases there is a HCl present and a gas is produced can you venture a guess what the gas can be?
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Rutherford

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Re: Problem of the week - 15/10/2012
« Reply #9 on: October 20, 2012, 02:20:06 PM »
If it's not an inconvenience, could you explain your beginning approach when tackling this problem? I'm interested to hear about it
Here is the whole approach. First I saw what Borek said, then I started from the last equation and then I went to the first. A is obviously a dichromate and F is a chloride of the metal in the dichromate. C is a chlorate of the metal, and I need only to put in the 31.83%. From the beginning I was thinking about potassium as the metal and then the calculation showed that it is surely potassium (I was thinking about H as water, because it is a common product of redox reactions in acidic solutions and it is present in every reaction in the problem). The rest is easy then.