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Topic: Please help me with my formula for C2H5C6H5  (Read 9708 times)

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Offline hahahanna

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Please help me with my formula for C2H5C6H5
« on: July 09, 2009, 11:35:24 AM »
Hello! Im stuck with this formula:

I have balance the 2 first formulas:


formel 1:

C2H5C6H5 + 24 H2O --> 8HCO3- + 50H+ + 42e-


Formel 2:

SO42- + 9H+ + 8e-  --> HS- 4H2O

Then im going to add formula 1 with 2. So I multiply as follow:


formel 1: (C2H5C6H5 + 24 H2O --> 8HCO3- + 50H+ + 42e-)*4
formel 2: (SO42- + 9H+ + 8e-  --> HS- 4H2O) * 21


The results is:

4C2H5C6H5 + 96H2O + 21SO42- + 189 H+ -->
32HCO3- + 200H+ + 21HS- + 84 H2O


Is this correct? And I want to fix this formula, because Im going now to calculate mass for C2H5C6H5S later.
How do I short this formula down?


Offline Borek

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Re: Please help me with my formula for C2H5C6H5
« Reply #1 on: July 09, 2009, 11:47:58 AM »
Cancel H+. I doubt this reaction makes any sense, but technically speaking it is correctly balanced.
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Offline hahahanna

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Re: Please help me with my formula for C2H5C6H5
« Reply #2 on: July 09, 2009, 11:57:08 AM »
But the thing I wonder if I can, hmm, can i just remove H+?

Because i find another fomula:

Formula 1: (C6H6 + 18 H2O -->6HCO3- + 36H+ + 30e-)*2
Formual 2: (O2 + 4H+ + 4e- --> 2H2O) *15

Result: 15O2 +2C6H6 + 6H2O --> 12HCO3- + 12H+


And, here they have just taking away H+ but how did they do that?
Did the in formula 1 just take: 36H+ - 30e- and then they just got 12H+ in the result?
And thats why there is no H+ in de left lead in the result?

Do you understand what i mean??

Offline Borek

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Re: Please help me with my formula for C2H5C6H5
« Reply #3 on: July 09, 2009, 12:34:13 PM »
Whatever have not changed during reaction can be removed from the equation.

So if you have

4H2 + O2 -> 2H2 + 2H2O

2H2 is just a spectator, and the reaction is

2H2 + O2 -> 2H2O
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Offline hahahanna

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Re: Please help me with my formula for C2H5C6H5
« Reply #4 on: July 09, 2009, 12:41:10 PM »
okej, but then the reaction must be as follow:

4C2H5C6H5 + 12H2O + SO42- --> 8HCO3- + 21HS- + 84 H2O

But can i remove H2O also? If i go along the theory you just showed?

Offline Borek

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Re: Please help me with my formula for C2H5C6H5
« Reply #5 on: July 09, 2009, 01:33:11 PM »
4C2H5C6H5 + 12H2O + SO42- --> 8HCO3- + 21HS- + 84 H2O

No, you can subtract exactly the same amount from both sides - not all, regardless of the stoichiometric coefficients.

Quote
But can i remove H2O also? If i go along the theory you just showed?

Yes, missed that.
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Offline hahahanna

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Re: Please help me with my formula for C2H5C6H5
« Reply #6 on: July 09, 2009, 04:08:36 PM »
Ok, so the reaction must be:


4C2H5C6H5 + 12H2O + 21SO42- --> 32HCO3- + 21 HS- ??


Offline Borek

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Re: Please help me with my formula for C2H5C6H5
« Reply #7 on: July 09, 2009, 05:42:40 PM »
Go back to H+.
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Offline hahahanna

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Re: Please help me with my formula for C2H5C6H5
« Reply #8 on: July 09, 2009, 06:03:52 PM »
4C2H5C6H5 + 84 H2O + SO42- --> 8HCO3- + 21HS-

Did i get it now??  :)

Offline Borek

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Re: Please help me with my formula for C2H5C6H5
« Reply #9 on: July 09, 2009, 06:47:35 PM »
No. You had 189H+ on the left and 200H+ on the right, how come there was nothing left on both sides?
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Offline hahahanna

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Re: Please help me with my formula for C2H5C6H5
« Reply #10 on: July 10, 2009, 05:42:59 AM »
4C2H5C6H5 + 12H2O + 21SO42- --> 32HCO3- + 21 HS- + 11H+

Can this be correct?

Offline AWK

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Re: Please help me with my formula for C2H5C6H5
« Reply #11 on: July 10, 2009, 06:25:51 AM »
AWK

Offline hahahanna

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Re: Please help me with my formula for C2H5C6H5
« Reply #12 on: July 10, 2009, 06:37:47 AM »
Yeah, the formula is correct but i want to short the formula down. If i use the webpage you gave me it will give me more in the fomula to short down.

Offline hahahanna

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Re: Please help me with my formula for C2H5C6H5
« Reply #13 on: July 10, 2009, 12:34:07 PM »
4C2H5C6H5 + 12H2O + 21SO42- --> 32HCO3- + 21 HS- + 11H+

Can this be correct?

Offline sjb

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Re: Please help me with my formula for C2H5C6H5
« Reply #14 on: July 10, 2009, 01:30:12 PM »
Well, as Borek has already said, it makes mathematical sense, but whether it makes chemical sense is another question. Certainly I've dried alkylbenzenenes over MgSO4 in organic solvents with a trace of water in and no such decomposition has occured, so if that means this is incorrect, then so be it.

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