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Topic: steamhydrocarbon reforming process  (Read 12272 times)

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spunkee

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steamhydrocarbon reforming process
« on: February 25, 2005, 11:29:21 PM »
The name is just scary enough... but here is my problem:

"How many kg of methane would be needed to produce 1 kg of hydrogen using the steamhydrocarbon reforming process? (Assume each step is 100% efficient.)"

The steamhydrocarbon reforming process is a three step reaction:

1. First Step At 1100 degrees C and a Ni catalyst:

H2O (g) + CH4 (g) --> CO (g) + 3H2 (g)

2. Second Step At 400 degrees C and metal oxide catalyst:

CO (g) + H2O (l) --> CO2 (g) + H2 (g)

3. No special temps or catalyst for this one:

CO2 (g) + 2OH[-] (aq) --> CO3[2-] (aq) + H2O (l)

Now, I am really lost on this one. Is this a molality problem? And if so, do I use all three reactions, just the first two, or just the first one. I tried finding a similar example of a problem like this in my book but it has been unsuccessful. The section that talks about this in my book simply gives the three reactions without explaining how to do a problem using them. All help is very much appreciated.  :animatedwink:

Offline Mitch

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Re:steamhydrocarbon reforming process
« Reply #1 on: February 25, 2005, 11:59:53 PM »
Since your only interested in Hydrogen as a product, you tell us which equations you should use. ;)
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spunkee

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Re:steamhydrocarbon reforming process
« Reply #2 on: February 26, 2005, 12:10:03 AM »
hmm, ok. I guess I am using more than one step because the wording in the question says "Assume each step is 100% efficient." This was what was confusing me initially but since both the first and second reactions produce hydrogen as a product, I am guessing I use the first two reactions?? (I'm sorry I am no good at chemistry, please don't laugh)

The tricky thing is, I just don't recall ever using two equations to solve a problem like this before and I can't find an example like it in my book.

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Re:steamhydrocarbon reforming process
« Reply #3 on: February 26, 2005, 12:15:55 AM »
You will only need equation 1 and 2. First calculate how many moles of hydrogen is 1 kg of hydrogen. In chemistry it is a lot easier to deal with things in moles not killograms.

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I'm sorry I am no good at chemistry, please don't laugh
We're here to help not to laugh.  :D
« Last Edit: February 26, 2005, 12:17:01 AM by Mitch »
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spunkee

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Re:steamhydrocarbon reforming process
« Reply #4 on: February 26, 2005, 12:24:05 AM »
ok there are 500 mol of H2 in one kilogram of H2

spunkee

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Re:steamhydrocarbon reforming process
« Reply #5 on: February 26, 2005, 12:25:49 AM »
We're here to help not to laugh.  :D

and it is MUCH appreciated. I try so hard to get chemistry and it just doesn't come easy for me most of the time :-[ so you are a blessing.

Ok, I think I know what to do. Use mole ratios to get the mole of methane and then convert that to kg? I know the first mole ratio is going to be 1 to 3 but I am unsure about the other mole ratio.
« Last Edit: February 26, 2005, 12:32:08 AM by spunkee »

dexangeles

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Re:steamhydrocarbon reforming process
« Reply #6 on: February 26, 2005, 04:43:49 AM »
hey Mitch,

why are both equations 1 and 2 needed?
i guess i've forgotten my gen chem  ???

Offline Mitch

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Re:steamhydrocarbon reforming process
« Reply #7 on: February 26, 2005, 04:48:28 AM »
Because the 2nd equation also produces H2 and it is made by one of the products of equation 1.
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spunkee

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Re:steamhydrocarbon reforming process
« Reply #8 on: February 26, 2005, 05:26:38 AM »
do you happen to know what the second mole ratio will be by any chance?  :penguinsmilie:

Offline Mitch

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Re:steamhydrocarbon reforming process
« Reply #9 on: February 26, 2005, 02:20:31 PM »
You to tell me how many moles of Hydrogen is in 1 kg of H2.
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spunkee

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Re:steamhydrocarbon reforming process
« Reply #10 on: February 26, 2005, 10:23:58 PM »
ok there are 500 mol of H2 in one kilogram of H2

I thought I figured that out already.. but maybe it is wrong  :-\? This is what I did to get it:
(1 kg H2) * (1000 g H2/1 kg H2) * (1 mol H2/2 g H2)= 500 mol/kg

OK, I think I figured this out... (as long as 1 kg of H2 has 500 moles, that is)
I took the 500 mol H2/kg and multiplied it by the mole ratios. 1 mole of H2 per 1 mole of CO and then 1 mol of CO per 1 mol of CH4, giving me 500 mol of CH4. Next I took 500 mol of CH4 and converted it to kg which gave me 8 kg CH4, and hopefully that is it...  :happy:
 :happy:
« Last Edit: February 26, 2005, 11:07:43 PM by spunkee »

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Re:steamhydrocarbon reforming process
« Reply #11 on: February 26, 2005, 11:16:52 PM »
Sorry you did do it right. I just missed that first post.

Now, I will do something very uncommon and give you a big chunck of the puzzle. This is just because it would be really hard to pepper you with questions, so that you could see it. I think in this case it works out better if I tell you the mole-to-mole relationship and you try to figure out why it is.

For every mole of CH4 you will produce 4 moles of H2.
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spunkee

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Re:steamhydrocarbon reforming process
« Reply #12 on: February 27, 2005, 07:43:49 PM »
hmm, I think that I MIGHT know this. I would say the mole ratio is 4 moles of Hydrogen for every one mole of methane because the initial reactant in equation 1 is one mole of CH4. This one mole yields 3 moles of H2 in the products of the first reaction and 1 mole of H2 in the product of the second reaction. If you add the total nuimber of moles of H2 yielded from both reactions, you get 4 moles, thus 1 mole of CH4 yields 4 moles TOTAL of H2. Hopefully, I'm right??  ???

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Re:steamhydrocarbon reforming process
« Reply #13 on: February 27, 2005, 08:05:30 PM »
Very close, but methane isn't in the second equation so how could it make H2?

What methane does make is 1 mol of CO, and that 1 mol of CO can go on in equation 2 to make 1 mol of H2.
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Offline Donaldson Tan

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Re:steamhydrocarbon reforming process
« Reply #14 on: February 27, 2005, 08:21:05 PM »
alternatively, given 100% conversion, you can try deriving the overall equation for this system, ie.

H2O +  CH4 + 2OH--> CO32- + 4H2
« Last Edit: February 27, 2005, 08:23:57 PM by geodome »
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