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 Posted: Wed Nov 14, 2007 2:53 pm
So there's this problem in my physical science hw that I don't get at all because I was absent the day my teacher covered it and the test's tomorrow and WAH!
Ammonia is manufactured by the haber process.
N2 + 3H2 [insert arrow pointing right over an arrow pointing left here] 2NH3 + heat
This involves the reaction of nitrogen with hydrogen to form ammonia. What mass of nitrogen is needed to make 34 g of ammonia?
I don't need the answer, I just need someone to explain how to get to the answer in a way that I can understand.
T-T
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Posted: Wed Nov 14, 2007 3:34 pm
Kitty Krazy So there's this problem in my physical science hw that I don't get at all because I was absent the day my teacher covered it and the test's tomorrow and WAH! Ammonia is manufactured by the haber process.
N2 + 3H2 [insert arrow pointing right over an arrow pointing left here] 2NH3 + heat
This involves the reaction of nitrogen with hydrogen to form ammonia. What mass of nitrogen is needed to make 34 g of ammonia?I don't need the answer, I just need someone to explain how to get to the answer in a way that I can understand. T-T hmmm, it's been a while since I have done much chemistry, but it seems to me that it is a problem of proportions. so it looks like you have a ratio of 1 N 2 to 3 H 2 to 2NH 3You're looking for nitrogen, which is 1 N 2 to 2NH 3, so set up a proportion. 34g NH 3 * (1 N 2 / 2NH 3) and then the ammonia "cancels out" and you are left with nitrogen. I hope this is really what you need. I can't remember if there is something I am forgetting or what (like does the fact that it is made by the haber process change anything), so you might want to get another opinion neutral
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Posted: Wed Nov 14, 2007 3:40 pm
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Posted: Wed Nov 14, 2007 4:01 pm
I MAY be wrong... so you should wait for Sibby "The master of everything mathematical/scientifical" to get here.
I think you multiply both sides by 34 so the equation looks kinda like:
34N2 + 102H2 ===> 68NH3 + heat
because "technically" you're measuring everything in grams. And normally the equation yields 2 grams on NH3. You only divide the big numbers. So it would look like:
17N2 + 51H2 ===> 34NH3 + heat
I BELIEVE that sounds right... sweatdrop
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Posted: Wed Nov 14, 2007 4:24 pm
Draconissa Kitty Krazy So there's this problem in my physical science hw that I don't get at all because I was absent the day my teacher covered it and the test's tomorrow and WAH! Ammonia is manufactured by the haber process.
N2 + 3H2 [insert arrow pointing right over an arrow pointing left here] 2NH3 + heat
This involves the reaction of nitrogen with hydrogen to form ammonia. What mass of nitrogen is needed to make 34 g of ammonia?I don't need the answer, I just need someone to explain how to get to the answer in a way that I can understand. T-T hmmm, it's been a while since I have done much chemistry, but it seems to me that it is a problem of proportions. so it looks like you have a ratio of 1 N 2 to 3 H 2 to 2NH 3You're looking for nitrogen, which is 1 N 2 to 2NH 3, so set up a proportion. 34g NH 3 * (1 N 2 / 2NH 3) and then the ammonia "cancels out" and you are left with nitrogen. I hope this is really what you need. I can't remember if there is something I am forgetting or what (like does the fact that it is made by the haber process change anything), so you might want to get another opinion neutral Your equation looks familiar to me (the science teacher did show me an example, but he took it away from me DX), but I don't understand it D: Why do you take the steps you take? I don't get why it works this way DX @Buzz - I don't think that's right... it doesn't look familiar at all o.o
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Posted: Wed Nov 14, 2007 4:34 pm
Kitty Krazy Draconissa Kitty Krazy So there's this problem in my physical science hw that I don't get at all because I was absent the day my teacher covered it and the test's tomorrow and WAH! Ammonia is manufactured by the haber process.
N2 + 3H2 [insert arrow pointing right over an arrow pointing left here] 2NH3 + heat
This involves the reaction of nitrogen with hydrogen to form ammonia. What mass of nitrogen is needed to make 34 g of ammonia?I don't need the answer, I just need someone to explain how to get to the answer in a way that I can understand. T-T hmmm, it's been a while since I have done much chemistry, but it seems to me that it is a problem of proportions. so it looks like you have a ratio of 1 N 2 to 3 H 2 to 2NH 3You're looking for nitrogen, which is 1 N 2 to 2NH 3, so set up a proportion. 34g NH 3 * (1 N 2 / 2NH 3) and then the ammonia "cancels out" and you are left with nitrogen. I hope this is really what you need. I can't remember if there is something I am forgetting or what (like does the fact that it is made by the haber process change anything), so you might want to get another opinion neutral Your equation looks familiar to me (the science teacher did show me an example, but he took it away from me DX), but I don't understand it D: Why do you take the steps you take? I don't get why it works this way DX @Buzz - I don't think that's right... it doesn't look familiar at all o.o OHHH... Yeah... well... my equations are never what the teachers use... I usually use the equation they have... exceot switched around to make it seem easier for me.
BUT Iin my defense, I said it might be wrong >.>
SIIIIIIIBBBBB WHERE ART THOU SIB!!!!!!! xd
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Posted: Wed Nov 14, 2007 5:00 pm
So, you have 34g of Ammonia. You know that the ratio between Ammonia and Nitrogen is 2 to 1, respectively.
This is the mole ratio. The, find the number of moles of Ammonia that you have.
This will be, about, 2, you do this by dividing the mass you have by the mass of one mole of the substance, which is the atomic masses of all the elements in the compound added together.
Then, use your mole ratio to find the number of moles of Nitrogen you need.
From that number, about 4, you multiply that number of moles by the molar mass of Nitrogen, which is about 28. That should give you the correct answer.
P.S. Refer to your own periodic table for more accurate masses than my general ones.
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Posted: Wed Nov 14, 2007 7:52 pm
What Azrel says makes sense, in that you have to do it with mols, not grams, because chemical equations deal with mols, not grams... but I can't say for sure... Kitty Krazy Draconissa Kitty Krazy Ammonia is manufactured by the haber process.
N2 + 3H2 [insert arrow pointing right over an arrow pointing left here] 2NH3 + heat
This involves the reaction of nitrogen with hydrogen to form ammonia. What mass of nitrogen is needed to make 34 g of ammonia? hmmm, it's been a while since I have done much chemistry, but it seems to me that it is a problem of proportions. so it looks like you have a ratio of 1 N 2 to 3 H 2 to 2NH 3You're looking for nitrogen, which is 1 N 2 to 2NH 3, so set up a proportion. 34g NH 3 * (1 N 2 / 2NH 3) and then the ammonia "cancels out" and you are left with nitrogen. I hope this is really what you need. I can't remember if there is something I am forgetting or what (like does the fact that it is made by the haber process change anything), so you might want to get another opinion neutral Your equation looks familiar to me (the science teacher did show me an example, but he took it away from me DX), but I don't understand it D: Why do you take the steps you take? I don't get why it works this way DX @Buzz - I don't think that's right... it doesn't look familiar at all o.o What Buzz did is a lot like what I did, but in a more directly-modifying kind of way. So, I'll try and explain, even though I didn't deal with the mols like Azrael did, which is probably more correct. Ok, so you have a ratio imbedded in the equation, right? 1 nitrogen + 3 hydrogen makes 2 ammonia + heat , so to compare nitrogen and ammonia, you take the numbers and create a ratio (aka devision) 1 N 2 / 2NH 3. you can use this with either one on top or bottom (aka 2NH 3/1N 2 if you need to) so now we have a ratio (a.k.a. something-to-use-to-convert). Then we take what you ended up with (the ammonium in this problem) (a.k.a. what-you-want-to-convert- from (you want to convert the ammonium you know about into the right amount of nitrogen) ) and apply the ratio to it. (1 N2) *34g NH 3 (2NH 3) = (1 N2) *34g NH3 (2 NH3) = (1) *34g N 2 --(2) That's what I was saying. But I did neglect to account for mols, so it may be incorrect, at least it isn't quite right. The process is the same, but you have to add the steps of converting to mols for the ammonium and then back to grams at the end for the nitrogen. That's kinda what Azrael was saying. I agree with Buzz that we need Sib to explain better and more correctly.
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Posted: Wed Nov 14, 2007 8:01 pm
I wish I remembered how to do that problem. I used to be good at it. >< Sib needs to come in with his present super knowledge of basic chem. >< I found this, hope it helps some. Grams to moles
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Posted: Wed Nov 14, 2007 8:31 pm
Okay, Okay... I'm here... ^_^( )
Well, for two ammonia molecules, you need one nitrogen molecule... Meaning that you need one moles of nitrogen for two moles of ammonia. Using the periodic table to figure out the atomic weights (average) of the elements involved ( hydrogen = 1, nitrogen = 14), then adding the sum of the weights gives you the number of grams per mole of substance. Meaning that NH3 = 14 + 1 +1 + 1 = 17 grams/mol
You have 34 g of NH3... so Divide 34g by 17 g/mol to get 2 moles of NH3... meaning that you need exactly 1 mole of Nitrogen... N2 has 28g/mole (14 + 14), and since you need exactly one mole of Nitrogen, you need 28 grams of N2 to make 34 g of NH3
So, it's pretty much what Azreal said... just a bit more longwinded... and hopefully a bit more revealing about what you need to do? o_o( )
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