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Stoichiometry problem help

  • Thread starter NEILS BOHR
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Homework Statement


A neutral gas sample contains 84% ( by volume ) of CH4 , 10 % of C2H6 , 3 % of C3H8 and 3% of N2 . If a series of catalytic reacns cud be used for converting all the C atoms of the gas into butadiene , C4H6 , with 100% efficiency , how much butadiene cud be prepared frm 100 g of the natural gas?


Homework Equations





The Attempt at a Solution


its an easy 1 , still not gettin the ans........plzz give some necessary hints
i think i m doin some mistake in % by volume thing...
 

Answers and Replies

  • #2
Borek
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How can we help you find the error not knowing what you did? Show your work.
 
  • #3
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well i calculated the % by volume of C in each compound............
converted into % by wt
then by dividing it by total wt
found the %age....
 
  • #4
Borek
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well i calculated the % by volume of C in each compound
It already doesn't make sense. You can't calculate % by volume of C in a compound.

Volumes can be used to calculate molar ratio of gases, that's application of the Avogadro's law.
 
  • #5
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so how to start??:confused:
 
  • #6
Borek
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I already told you - start calculating molar ratios.
 
  • #7
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hmmm

if 1 mole CH4 contains 1 mole C

how can we find moles of C in 10 % CH4...??
 
  • #8
Borek
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Not 10% of CH4, but 10% of the mixture.

Imagine you have 100L of the mixture - can you use Avogadro's law to calculate its mass?
 
  • #9
symbolipoint
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Not 10% of CH4, but 10% of the mixture.

Imagine you have 100L of the mixture - can you use Avogadro's law to calculate its mass?
He may be able to calculate his solution without using Avogadro's Number. For an ideal gas, gas is 22.4 Liters per Mole. This makes me think that gas percent compounds by volume is the same as mole percent. From number of moles of a given gas sample volume, the mass of each compound can be caclulated (for which the molecular weights of each compound is needed, first).

With that mass information of each compound, now reference can be made to 100 grams sample of the gas. Mass percents of each compound can give mass in the 100 gram sample, and moles of each compound in the 100 gram sample can be found.

NOW, refer to balanced reaction.

NOTE: Misinterpreted Avogadro's Number with Avogadro's Law.
 
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  • #10
Borek
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This makes me think that gas percent compounds by volume is the same as mole percent.
Good, you just applied http://en.wikipedia.org/wiki/Avogadro's_law :smile:

Avogadro's law is not about Avogadro's number. It is about about same volumes of different gases containing the same number of molecules, hence the same number of moles.
 
  • #11
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so you mean i should find number of moles by taking the help of given % by volume...
 
  • #12
symbolipoint
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I tried working through the problem myself several hours ago. What I find troubling is that I could not find a balanced reaction. What could I possibly be missing? Could I ignore the presence of Nitrogen gas or does it participate in a reaction? I see no clear accounting for the numbers of atoms nor other adjustments; only the mole fractions present in the given information. I had usually done simple combustion reaction balancing well, and redox types well for balancing, but on the one under discussion here, I'm lost. "All of the C is converted to C4H6"..., then I'm not sure how to account for C or H. Is H2 produced?
 
  • #13
Borek
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You are making it harder than it is. How many moles of C? How many moles of C per mole of butadiene?
 
  • #14
symbolipoint
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You are making it harder than it is. How many moles of C? How many moles of C per mole of butadiene?
So the exercise does not really require a balanced equation; only accounting for the number of Carbon atoms? Does this mean that the stoichiometry comes directly from the mole fractions of the organic gases?
 
  • #15
Borek
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So the exercise does not really require a balanced equation; only accounting for the number of Carbon atoms?
That's how I understand it.

Does this mean that the stoichiometry comes directly from the mole fractions of the organic gases?
Can you elaborate? I am not sure what you are asking about.
 
  • #16
symbolipoint
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That's how I understand it.



Can you elaborate? I am not sure what you are asking about.
About posts #14 and 15, I feel the need to form a balanced equation, but the example problem does not give any hints at what may be possible. The gas sample has mole fractions for each reactant compound. The product is butadiene, C4H6. I find no further way to account for the C on both sides and the H on both sides.

I do not find possible stoichiometry for this exercise in the way that I had learned and used in school; I had been accustomed to balancing reactants and products regarding the moles of each atom or element. In the butadience from the natural gas conversion here, the only possibility is to use all of the carbons present in the 100 gram sample, and determine (calculate) the equivalent amount of butadiene. This is my best understanding.
 
  • #17
Borek
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OK, I think I know what you mean.

The way I see it it is just a question on mass conservation.

Note that often you don't need reaction equation to calculate the result. For example if you are told Fe3O4 was converted in a series of reactions to a Prussian blue (Fe7(CN)18⋅14H2O), you don't need to know exact reactions and exact stoichiometry of each - just from the iron balance it is obvious that one mole of Fe3O4 can yield 3/7 moles of Fe7(CN)18⋅14H2O (probably less if the yields are not 100% - but you have an obvious upper limit).
 
  • #18
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i dont know why i am getting stuck in such a question???

can u explain it in a little easier way??:redface:
 
  • #19
symbolipoint
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I solved it a few hours ago. Yes, I strictly used mass conversion based on the equivalent amount of Carbon from the 100 gram sample of gas. All Carbon becomes its equivalent amount of butadiene, 100%.

My solution process used a table of all the four gasses and I performed successive calculations for each one.

I started with reference to 1 liter of gas, then computed number of moles of each compound based on 22.4 Liter per Mole of molecules. From this, I computed amount of grams of each compound present in this 1 liter of gas (meaning "natural gas sample"). I DID use molecular weight of each compound. Sum them, and I have how many GRAMS of the 1 Liter natural gas.
... long drawn-out process continues, but putting into a table made more sense.
... My answer was... could I give it on the forum?
 
  • #20
morrobay
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Yes I also went through that long drawn out process from liters to moles to grams C
But it did not seem like enough grams carbon given the sample size.
I think the following is valid, if unconventional:
Suppose the total 100 g in three cases of each gas was 100% of that particular gas.
100g CH4 = 75g C
100g C2H6 = 80g C
100g C3H8 =81.8 g C
Now take the percentages of those gases actually present: add up the grams carbon.
I got xxx moles C4H6
 
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  • #21
symbolipoint
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morrobay, so your are finding 24 grams of C4H6.

Edit: no, that's not what he said. I think you used wrong molar mass, as morrobay number of moles was correct.
 
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  • #22
morrobay
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Edit by Borek: please beware, OP still have not solved the question on his own. I think a week should be enough waiting, we can continue the discussion on Monday.

You both got the same answer. That's the same answer I got.
 
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  • #23
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@symbolipoint

what is the mass of the natural gas that you are getting??

by your method , i am not getting the right answer!!
 
  • #24
Borek
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what is the mass of the natural gas that you are getting??

by your method , i am not getting the right answer!!
Instead of asking for our results, show your calculations. Just knowing the final number won't help you to understand how it was calculated, nor it will help us help you find the mistake.
 
  • #25
symbolipoint
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About posts #23 and #24,
My descriptions of what I did are maybe not very easy to understand and they may need refinement. Actually setting up the calculations and arranging a table were easier than trying to explain what I did. Still, much careful units and formula steps.
 

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