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Offline nupatel87

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Excess Reagent problem
« on: September 30, 2009, 12:41:07 AM »
Hello everyone, I found these forums while searching for guidance for this question.
A mixture of 4.52 g of H2 and 24.80 g O2 reacts to form water. The substances present when the reaction is over are H2O and H2.  27.12g of H2O are formed. 

How many grams of excess reagent remain after this reaction is complete.

I know that you are supposed to set up a mole ratio, but no matter what significant figures I used, I still can't get an answer.

Any help would be appreciated.

Offline AWK

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Re: Excess Reagent problem
« Reply #1 on: September 30, 2009, 03:18:14 AM »
4.52 + 24.80 = 27.12 + x
AWK

Offline Borek

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Re: Excess Reagent problem
« Reply #2 on: September 30, 2009, 03:21:06 AM »
Show what and how you did.

AWK approach is much faster and is based directly on the mass balance, but it can't be used blindly and in each case. Here it fits perfectly :)
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Offline nupatel87

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Re: Excess Reagent problem
« Reply #3 on: September 30, 2009, 08:21:54 AM »
Oh sorry I didn't post my work.
Well first of all we are given that the chemical equation ends up like this :

3H2 +O2 ---> 2H2O + H2
We are given the mass of H2 initially is 4.52g and the mass of O2 initially is 24.8g. When I use mole ratios I got the correct ending result of H20 to be 27.12g. When I converted both H2 and O2 to H2O, the mass of H2O produced by H2 was 27.12g and the mass of H2O produced by O2 was 27.9g. This means that the O2 was the excess reagent.

So when I entered those parts of the question in, I got them right, but when it asked for the amount of excess reagent in grams remaining, I can't seem to find the appropriate answer. I know that when doing excess reagent calculation, you take the amount of product that was produce, in this case 27.12g H2O, and then convert it back to the reagent in question(O2). So when I set up a mole ratio it ends up:

(27.12gH2O) x (1 mol H2O)/(18.016 g H2O)  x (1 mol O2)/(2 mol H2O) x (32 g O2)/(1 mol O2) = __ g O2 used
Thus 24.8g O2 - __g O2 used = mass of remaining excess.

When I did this I got the answer to be .7147g of excess O2, which works out mathematically if trying to convert it back to grams of H2O. I messed around with the 27.12 and changed it to more exact numbers so I also got .880g O2 and .693g O2.

Is this work the work you wanted me to show?

 

Offline nupatel87

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Re: Excess Reagent problem
« Reply #4 on: September 30, 2009, 09:47:13 AM »
I tried the method posted by AWK, and I got 2.2g, but when I entered it in, it was wrong :(

Offline DrCMS

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Re: Excess Reagent problem
« Reply #5 on: September 30, 2009, 10:10:09 AM »
The person who set the question this way is trying to be very sneaky but is actually too stupid to pull it off correctly.

The reaction between hydrogen and oxygen to produce water is:

2H2 +O2 ---> 2H2O

which given the start weights means hydrogen is in excess and 27.9g water would be produced with 1.42g hydrogen left.

but you have been given:

3H2 +O2 ---> 2H2O+ H2

which suggest the oxygen is in excess giving 27.12g water plus 0.69g oxygen and 1.51g hydrogen.  However, any hydrogen produced in this reaction would react with the oxygen left over to give more water.

AWK's answer is correct for the actual chemical reaction involved but wrong for the stupid made up one you have been given.   I suggest you get your teacher to contact us so we can point out what an idiot they are.




Offline Borek

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Re: Excess Reagent problem
« Reply #6 on: September 30, 2009, 10:16:08 AM »
First of all - are you sure it was 27.12 and not 27.92?

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Offline Borek

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Re: Excess Reagent problem
« Reply #7 on: September 30, 2009, 10:17:16 AM »
The person who set the question this way is trying to be very sneaky but is actually too stupid to pull it off correctly.

Or made a simple stupid typo. It happens.
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Offline DrCMS

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Re: Excess Reagent problem
« Reply #8 on: September 30, 2009, 10:50:10 AM »
The person who set the question this way is trying to be very sneaky but is actually too stupid to pull it off correctly.

Or made a simple stupid typo. It happens.

And made another one when they gave the wrong reaction equation of :

Well first of all we are given that the chemical equation ends up like this :

3H2 +O2 ---> 2H2O + H2


But if you use this wrong reaction equation you get 27.12g of water produced from a single pass with the weights of raw materials given.  So I think it very unlikely this was a typo.  I think the teacher was attempting to be sneaky but failed because they're too stupid to do it correctly.

Offline Borek

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Re: Excess Reagent problem
« Reply #9 on: September 30, 2009, 12:06:57 PM »
Judging from the fact that nupatel was not able to locate the problem, it was sneaky enough.
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Offline DrCMS

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Re: Excess Reagent problem
« Reply #10 on: September 30, 2009, 02:55:32 PM »
Yes it was sneaky but it is also stupid dumb halfwited rubbish, teachers should be sacked for being this stupid in my opinion.

Offline nupatel87

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Re: Excess Reagent problem
« Reply #11 on: September 30, 2009, 03:03:28 PM »
Thanks for the help I really appreciate it ;D. That H2 in the products really threw me off. :P

Offline DrCMS

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Re: Excess Reagent problem
« Reply #12 on: September 30, 2009, 03:27:29 PM »
That H2 in the products really threw me off. :P

It would do as it is unscientific rubbish.  A material can not be a reagent and a product of a single reaction it is just plain wrong.  The person who set the question is an idiot if they do not explain it was made up just to test your thinking.

Offline AWK

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Re: Excess Reagent problem
« Reply #13 on: October 01, 2009, 02:12:48 AM »
I tried the method posted by AWK, and I got 2.2g, but when I entered it in, it was wrong :(
3 significant digits are needed
AWK

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