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Topic: Stoichiometric ratio and Stoichiometry; What is the difference?  (Read 17027 times)

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Offline Scorael

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Stoichiometric ratio and Stoichiometry; What is the difference?
« on: January 31, 2009, 12:01:33 AM »
Hi guys,
I'm an electrical engineering undergrad working on a hydrogen fuel cell simulation. I have been using a book to help me with this but there are some terms that I do not understand properly. I hope someone can help out.

In the book, the author mentioned the stoichimetry and the stoichiometric ratio of air in 2 separate examples. The stoichiometric ratio is was used to calculate oxygen flow rate while the stoichiometry is for air flow rate (I'm assuming they actually mean the same because both are listed as 2.5 but I'll rather get a 2nd opinion).
I need to know the difference between the stoichiometric ratio and stoichiometry and how do I obtain them.

Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #1 on: January 31, 2009, 06:02:34 AM »

Offline Scorael

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #2 on: February 01, 2009, 12:48:25 AM »
Hi Argos,
Those links did help my understanding of the terms. I now understand the significance of stoichiometry in the equation (turns out it meant the fuel-air ratio). There's still some confusion on the meaning of stoichiometric ratio and I hope you can help me further.

I am given these equations to calculate mass flow rate of Oxygen and Hydrogen into the cell:

mO2 = 0.5*SO2*MO2*I/(2F)
mH2 = SH2*MO2*I/(2F)

The flow of unused reactants out of the cell is given as:
mO2 = 0.5*(SO2-1)*MO2*I/(2F)
mH2 = (SO2-1)*MO2*I/(2F)

mi is the mass flow rate of species i in g/s
Si is termed as the stoichiometric ratio of species i
Mi is the molar mass of species i
F is the Faraday's number
I is the current drawn

What I don't really get is the stoichiometric ratio in here. From my understanding, the reaction of hydrogen and oxygen to form water is H2+0.5O2 = H2O giving oxygen the stoichiometric ratio of 0.5 but clearly in the outlet flow equation S can be bigger than 1 (In 1 example S = 2).

Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #3 on: February 01, 2009, 11:12:29 AM »

Offline Scorael

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #4 on: February 03, 2009, 06:27:30 AM »
I don't have a scanner so I can't really do that.

Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #5 on: February 03, 2009, 08:47:24 AM »

Offline Scorael

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #6 on: February 03, 2009, 09:41:07 AM »
After studying the earlier chapters I realise the equation is I/(nF), n being the number of moles of electron, which appears to be what you have written.

Still does not explain where did stoichiometric ratio, S, came from though. I am given S in the question as 2 but I am not sure whether they arrived at that value through measurements or is it a constant.

I'll update later. Need to sleep now. Thanks for all the help though.

Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #7 on: February 03, 2009, 11:52:20 AM »

Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #8 on: April 29, 2009, 04:13:25 PM »
                            Copy of lost replay No. 1

Dear Scorael;

Stoichiometric ratio is an absolutely required part of the whole stoichiometry.
Stoichiometric ratio is dealing with moles whereas stoichiometry deals with both moles and masses.
For Stoichiometric ratio you may best visit Mr. Borek's:   

For doing the whole stoichiometry you may then follow the simple recipe on:
Therein is also an Example Diagram: How to do a “Stoichiometry Problem”.

I hope to have been of help to you.
Good Luck!
                    ARGOS++


Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #9 on: April 29, 2009, 04:21:05 PM »
                             Copy of lost replay No. 3

Dear Scorael;

After several tries to interpret these equations as a model, do you allow that I say it in plain language?:
They don’t make real sense to me at all; and that for several reasons!

I have Questions, three times the number of equations you have given!

Are you able to scan the few pages around your problem and send them to me via pm?
I hope this way I will have more chances to interpret the model correctly.

Sorry!, - but I think that’s the only chance we will have.
Good Luck!
                    ARGOS++

Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #10 on: April 29, 2009, 04:26:36 PM »
                             Copy of lost replay No. 5

Dear Scorael;

In this moment I can give you only what I did for the model:

I used the following Equations:
   A.)   Ox:       O    +  2e-    ---->    O2-
   B.)   Red:      H                ---->    H+    +  1e-
   C.)   RxN:     2H2  +  1O2   ---->   2H2O
   D.)   Current :    moles e- / sec  =  I / F =  I * 1 C/sec /  96’485.3 C/mole

So I got for the mass flow(in) as a function of the current:
mO2(in) =  I / F  / noxO  / rxnO  * MWO2 = I / 96’485.3 / 2 / 2 * 32.0  g/mole = x mole O2 / sec.
mH2(in) =  I / F  / nredH  / rxnH * MWH2 =  I / 96’485.3 / 1 / 2 *  2.0  g/mole = x mole H2 / sec.
With:
     -    noxO   =   2 from A.)    and  rxnO =  1 * 2 from C.)
     -    nredH  =   1 from B.)    and  rxnH =  2 * 1 from C.)

Or simpler spoken:
If you combine A.) with C.) or B.) with C.) it results that with one RxN you have a generation/exchange of 4e- (respective 4 moles) and for that 1.0 mole O2 and 2 moles H2 is required.
That’s all valid for a conversion efficiency of 100%.

You may think about and tell me your conclutions.

I hope to have been of help to you.
Good Luck!
                    ARGOS++

Offline ARGOS++

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Re: Stoichiometric ratio and Stoichiometry; What is the difference?
« Reply #11 on: April 29, 2009, 04:29:09 PM »
                             Copy of the lost replay No. 7

Dear Scorael;

Sorry!, -  My old mistake, because I very rarly have to deal with RedOx:
    I interchanged the terms  Red and Ox once again!

Here the corrected Equations:
   A.)   Red:     O    +  2e-    ---->    O2-
   B.)   Ox:       H                ---->    H+    +  1e-
   C.)   RxN:     2H2  +  1O2   ---->   2H2O
   D.)   Current :    moles e- / sec  =  I / F =  I * 1 C/sec /  96’485.3 C/mole

And:
mO2(in) =  I / F  / nredO  / rxnO  * MWO2 = I / 96’485.3 / 2 / 2 * 32.0  g/mole = x mole O2 / sec.
mH2(in) =  I / F  / noxH   / rxnH   * MWH2 =  I / 96’485.3 / 1 / 2 *  2.0  g/mole = y mole H2 / sec.
With:
     -    nredO  =   2 from A.)    and  rxnO =  1 * 2 from C.)
     -    noxH   =   1 from B.)     and  rxnH =  2 * 1 from C.)


Be remembered that it is bad practice to use the term: I /(nF), because I/F is clearly indicating the current in mole electrons per second, as pointed out in equation D.)!

The stoichiometric ratio in your case is = mole O2  / mole H2   = x / y = 1/2 (from above, but it follows also from C.)).
But then using 0.5 and Sx and (nF) in the same equation doesn’t make any sense, as in your equations.

Good Luck!
                    ARGOS++

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