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Author Topic: Help With A Chemistry Problem (Limiting Reagents/Reactants)?  (Read 3050 times) Average Rating: 0
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« on: April 21, 2012, 09:52:00 AM »

"For the reaction represented by 2Na + 2H2O ----> 2NaOH + H2, how many grams of hydrogen are produced if 120 grams of Na and 80 grams of H2O are available?"

The answer is "4.5 grams" (or so I was told, by a colleague of mine who had gotten the correct answer), but I was not sure how to derive this answer. This is the only question on the exam that I had gotten wrong and I was wondering if one of you guys knew how to solve it. Please excuse my ignorance!

Thanks and God bless!
« Last Edit: April 21, 2012, 10:10:07 AM by Severian » Logged

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« Reply #1 on: April 21, 2012, 10:31:48 AM »

"For the reaction represented by 2Na + 2H2O ----> 2NaOH + H2, how many grams of hydrogen are produced if 120 grams of Na and 80 grams of H2O are available?"

The answer is "4.5 grams" (or so I was told, by a colleague of mine who had gotten the correct answer), but I was not sure how to derive this answer. This is the only question on the exam that I had gotten wrong and I was wondering if one of you guys knew how to solve it. Please excuse my ignorance!

Thanks and God bless!

I get 4.44 g H2.
The mw of Na is 23 g/mole so 120 g gives you ~5.2 moles
The mw of H2O is 18 g/mole so 80 g gives you 4.444 moles. So H2O is limiting.
You are liberating 1 atom of H per H20 so that is 4.444 moles
The mw of H is 1 g/mole, so you get 4.444 grams of H2

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« Reply #2 on: April 21, 2012, 10:33:01 AM »

"For the reaction represented by 2Na + 2H2O ----> 2NaOH + H2, how many grams of hydrogen are produced if 120 grams of Na and 80 grams of H2O are available?"

The answer is "4.5 grams" (or so I was told, by a colleague of mine who had gotten the correct answer), but I was not sure how to derive this answer. This is the only question on the exam that I had gotten wrong and I was wondering if one of you guys knew how to solve it. Please excuse my ignorance!

Thanks and God bless!
1. First find the moles of each reactant: Na has 23 g/mol and H2O has 18; you get this from the periodic table. So divide the 120 grams of Na by 23g/mol and the 80 g of H2O by 18g/mol and you should get 5.17 Mol Na and 4.44 Mol H2O

2. Since the coefficient 2 is before each reactant, multiply the moles by 2 to get 10.34 mol Na and 8.88 mol H2O

3. Since there are less moles of H2O you know that this is the limiting reactant so only 8.88 moles of H2O and 8.88 moles of Na are being consumed in this reaction. The rest is excess.

4. Going through the reaction, for every two moles of Water and Sodium, only one mole of Hydrogen gas is being produced.
Therefore, 8.88 moles of Water and Sodium will result in 4.44 moles of Hydrogen gas.
Hydrogen gas has a atomic weight of 2.016 g/mol so theoretically the answer should be around 9 grams of H2.

Someone else can check my work but I'm pretty sure this is the right answer.
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« Reply #3 on: April 21, 2012, 10:34:11 AM »

"For the reaction represented by 2Na + 2H2O ----> 2NaOH + H2, how many grams of hydrogen are produced if 120 grams of Na and 80 grams of H2O are available?"

The answer is "4.5 grams" (or so I was told, by a colleague of mine who had gotten the correct answer), but I was not sure how to derive this answer. This is the only question on the exam that I had gotten wrong and I was wondering if one of you guys knew how to solve it. Please excuse my ignorance!

Thanks and God bless!

I get 4.44 g H2.
The mw of Na is 23 g/mole so 120 g gives you ~5.2 moles
The mw of H2O is 18 g/mole so 80 g gives you 4.444 moles. So H2O is limiting.
You are liberating 1 atom of H per H20 so that is 4.444 moles
The mw of H is 1 g/mole, so you get 4.444 grams of H2


It's not H being produced but H2 right? So shouldn't you have to double the grams?
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« Reply #4 on: April 21, 2012, 10:36:14 AM »

Well, just for everyone's reference; Hydrogen gas has to be written as "H2" as it is diatomic by/in nature. Just putting that out there in case no one knew that...

Thank you all for your contributions so far.
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« Reply #5 on: April 21, 2012, 11:28:38 AM »

It's not H being produced but H2 right? So shouldn't you have to double the grams?
Not AFAIK, because Hydrogen gas is diatomic by nature, so it is written as H2 "by default" (so to speak). I could be wrong, however. It's been a while since I've practiced limiting reagent/stoichiometry problems.
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« Reply #6 on: April 21, 2012, 11:55:37 AM »

"For the reaction represented by 2Na + 2H2O ----> 2NaOH + H2, how many grams of hydrogen are produced if 120 grams of Na and 80 grams of H2O are available?"

The answer is "4.5 grams" (or so I was told, by a colleague of mine who had gotten the correct answer), but I was not sure how to derive this answer. This is the only question on the exam that I had gotten wrong and I was wondering if one of you guys knew how to solve it. Please excuse my ignorance!

Thanks and God bless!

I get 4.44 g H2.
The mw of Na is 23 g/mole so 120 g gives you ~5.2 moles
The mw of H2O is 18 g/mole so 80 g gives you 4.444 moles. So H2O is limiting.
You are liberating 1 atom of H per H20 so that is 4.444 moles
The mw of H is 1 g/mole, so you get 4.444 grams of H2


It's not H being produced but H2 right? So shouldn't you have to double the grams?

Quick note. 4.444 moles of H released is 2.222 mole of H2 gas. The grams remains the same.
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« Reply #7 on: April 21, 2012, 11:59:28 AM »

dang guys, i'm impressed!  Cool
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« Reply #8 on: April 21, 2012, 12:18:26 PM »

It's been a while (12 years) since I've done this, but...

Here is the original reaction with weights
(120g)2Na + (80g)2H2O -> 2NaOH + H2

Converting the weights to moles and dividing by 2 to account for the coefficient of 2 (which I did not account for in step 1)
120/23=5.2174=2*2.6087        80/18=4.4444=2*2.2222

Reaction in moles including and accounting for coefficient
(2.6087m)2Na + (2.2222m)2H20 -> (?m)2NaOH + (?m)H2

And again accounting for the water being the limiting factor
(0.3865m)2Na + (2.2222m)2Na + (2.2222m)2H2O ->
(0.3865m)2Na + (2.2222m)2NaOH + (2.2222m)H2

Converting the moles back into weight
(8.8895g)2Na + (88.8888g)2NaOH + (4.4444g)H2

And finally multiplying by coefficients
(17.779g)Na + (177.7776g)NaOH + (4.4444g)H2

This should be what the reaction would yield if my (and most others here) math is correct.
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« Reply #9 on: April 21, 2012, 12:22:02 PM »

It's been a while (12 years) since I've done this, but...

Here is the original reaction with weights
(120g)2Na + (80g)2H2O -> 2NaOH + H2

Converting the weights to moles and dividing by 2 to account for the coefficient of 2 (which I did not account for in step 1)
120/23=5.2174=2*2.6087        80/18=4.4444=2*2.2222

Reaction in moles including and accounting for coefficient
(2.6087m)2Na + (2.2222m)2H20 -> (?m)2NaOH + (?m)H2

And again accounting for the water being the limiting factor
(0.3865m)2Na + (2.2222m)2Na + (2.2222m)2H2O ->
(0.3865m)2Na + (2.2222m)2NaOH + (2.2222m)H2

Converting the moles back into weight
(8.8895g)2Na + (88.8888g)2NaOH + (4.4444g)H2

And finally multiplying by coefficients
(17.779g)Na + (177.7776g)NaOH + (4.4444g)H2

This should be what the reaction would yield if my (and most others here) math is correct.



Verrry niiiice!
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« Reply #10 on: April 21, 2012, 01:29:42 PM »

"For the reaction represented by 2Na + 2H2O ----> 2NaOH + H2, how many grams of hydrogen are produced if 120 grams of Na and 80 grams of H2O are available?"

The answer is "4.5 grams" (or so I was told, by a colleague of mine who had gotten the correct answer), but I was not sure how to derive this answer. This is the only question on the exam that I had gotten wrong and I was wondering if one of you guys knew how to solve it. Please excuse my ignorance!

Thanks and God bless!
1. First find the moles of each reactant: Na has 23 g/mol and H2O has 18; you get this from the periodic table. So divide the 120 grams of Na by 23g/mol and the 80 g of H2O by 18g/mol and you should get 5.17 Mol Na and 4.44 Mol H2O

2. Since the coefficient 2 is before each reactant, multiply the moles by 2 to get 10.34 mol Na and 8.88 mol H2O

3. Since there are less moles of H2O you know that this is the limiting reactant so only 8.88 4.44 moles of H2O and 8.884.44 moles of Na are being consumed in this reaction. The rest is excess.

4. Going through the reaction, for every two moles of Water and Sodium, only one mole of Hydrogen gas is being produced.
Therefore, 8.884.44 moles of Water and Sodium will result in 4.442.22moles of Hydrogen gas.
Hydrogen gas has a atomic weight of 2.016 g/mol so theoretically the answer should be around 9 4.5 grams of H2.

Someone else can check my work but I'm pretty sure this is the right answer.
Yeah, I just checked my work and I'm an idiot. Nothing to see here folks; move along.
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« Reply #11 on: April 21, 2012, 01:50:24 PM »

The answer "4.44 g" seems to be correct. It is possible that the instructor rounded this answer to "4.5 g" for the sake of convenience, as he is not very strict in regards to sig figs and rounding numbers, etc. Thank you all for this most elucidating thread!
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« Reply #12 on: April 23, 2012, 04:09:17 PM »

Could someone perhaps give me a demonstration as to how to apply limiting reactants to gas stoichiometry? We will be testing on it sometime this week and while I have mastered stoichiometry I am still struggling a bit with limiting reactants. The concept is just not entering my brain fully, I suppose. Do forgive me if this request is too vague.
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« Reply #13 on: April 23, 2012, 05:07:31 PM »

Could someone perhaps give me a demonstration as to how to apply limiting reactants to gas stoichiometry? We will be testing on it sometime this week and while I have mastered stoichiometry I am still struggling a bit with limiting reactants. The concept is just not entering my brain fully, I suppose. Do forgive me if this request is too vague.

The limiting reactant is what will not have any left over after the reaction.

To use the example above (I know it's not gas but will get the point across), you have 5.2mol of Na and 4.4mol of H2O when the reaction calls for equal amounts of each. The 4.4mol of water will only react with 4.4mol of the Na leaving leftover Na. Becuase you use all the water and have leftover Na, the water is the limiting reactant in this case.

I hope that helps.
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« Reply #14 on: April 23, 2012, 05:29:27 PM »

^Thank you very much. Smiley
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« Reply #15 on: May 16, 2012, 07:58:24 PM »

Could someone be so kind as to walk me through a titration problem? I remember how to find pH/pOH, as those problems are pretty simple, but I forgot how to solve titration problems.

God bless and please pray for me and my family.
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« Reply #16 on: May 16, 2012, 08:18:46 PM »

Could someone be so kind as to walk me through a titration problem? I remember how to find pH/pOH, as those problems are pretty simple, but I forgot how to solve titration problems.

God bless and please pray for me and my family.


Can you give an example problem so I know what type of problem you are writing about? Are you trying to determine the concentration of an unknown with a known or is it something else?
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« Reply #17 on: May 16, 2012, 08:26:32 PM »

greetings in that divine and most precious name of our lord and savior jesus christ!
It's been a while (12 years) since I've done this, but...

Here is the original reaction with weights
(120g)2Na + (80g)2H2O -> 2NaOH + H2

Converting the weights to moles and dividing by 2 to account for the coefficient of 2 (which I did not account for in step 1)
120/23=5.2174=2*2.6087        80/18=4.4444=2*2.2222

Reaction in moles including and accounting for coefficient
(2.6087m)2Na + (2.2222m)2H20 -> (?m)2NaOH + (?m)H2

And again accounting for the water being the limiting factor
(0.3865m)2Na + (2.2222m)2Na + (2.2222m)2H2O ->
(0.3865m)2Na + (2.2222m)2NaOH + (2.2222m)H2

Converting the moles back into weight
(8.8895g)2Na + (88.8888g)2NaOH + (4.4444g)H2

And finally multiplying by coefficients
(17.779g)Na + (177.7776g)NaOH + (4.4444g)H2

This should be what the reaction would yield if my (and most others here) math is correct.


stay blessed,
habte selassie
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« Reply #18 on: May 16, 2012, 08:45:19 PM »

Could someone be so kind as to walk me through a titration problem? I remember how to find pH/pOH, as those problems are pretty simple, but I forgot how to solve titration problems.

God bless and please pray for me and my family.


Can you give an example problem so I know what type of problem you are writing about? Are you trying to determine the concentration of an unknown with a known or is it something else?
How about these...

For the following acid-base titration combination, determine the number of moles of the first substance listed that would be required to react with the given quantity of the second: NaOH with 1.0 mole HCL.

And:

If 15.0 mL of 0.0250 M aqueous H2SO4 is required to neutralize 10.0 mL of an aqueous solution of KOH, what is the molarity of the KOH solution?

Thanks again.
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« Reply #19 on: May 16, 2012, 09:51:31 PM »

For the following acid-base titration combination, determine the number of moles of the first substance listed that would be required to react with the given quantity of the second: NaOH with 1.0 mole HCL.

The objective is to have equal levels of H and OH. Because HCL has the same amount of H as NaOH has of OH, the would require the equal amounts to balance each other out. The answer would be 1.0 mole of NaOH.

Quote
And:If 15.0 mL of 0.0250 M aqueous H2SO4 is required to neutralize 10.0 mL of an aqueous solution of KOH, what is the molarity of the KOH solution?

.0750 if I remember the math correctly.
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« Reply #20 on: May 16, 2012, 10:14:49 PM »

Could someone be so kind as to walk me through a titration problem? I remember how to find pH/pOH, as those problems are pretty simple, but I forgot how to solve titration problems.

God bless and please pray for me and my family.


Can you give an example problem so I know what type of problem you are writing about? Are you trying to determine the concentration of an unknown with a known or is it something else?
How about these...

For the following acid-base titration combination, determine the number of moles of the first substance listed that would be required to react with the given quantity of the second: NaOH with 1.0 mole HCL.

And:

If 15.0 mL of 0.0250 M aqueous H2SO4 is required to neutralize 10.0 mL of an aqueous solution of KOH, what is the molarity of the KOH solution?

Thanks again.


Acid and base titrations of this type mean that they want the equivalent number of Acid moities, in the case H+ and Base moities OH-.

In the first problem  NaOH +  HCl --> NaCl + H20, so one of each. One mole of NaOH is necessary to titrate one mole of HCl.

The second question is posed as a simple algebra problem with the same principles.

The equation  H2SO4 + 2KOH --> K2SO4 + 2H2O

So it takes 2 KOH to neutralize 1 H2SO4. In other words here B (for base) must equal 2 A (the acid) in concentration. (B=2A)

15 ml X 0.025 M A = 10 ml X Q(unknown) M B
Since B=2A
15 ml x 0.025 M A = 10 ml X Q X 2 X M A

Q M = (15 ml x 0.025 M A) divided by (10 ml x 2 M A)

I hope this makes sense. M of course refers to molar as the unit specified in the problem.


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« Reply #21 on: May 20, 2012, 12:25:10 PM »

Thank you all for contributing. I will be sure to continue using this thread as a helpful chem reference until the sememester ends.
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« Reply #22 on: May 21, 2012, 12:38:08 PM »

Greetings in that Divine and Most Precious Name of Our Lord and Savior Jesus Christ!

Thank you all for contributing. I will be sure to continue using this thread as a helpful chem reference until the sememester ends.

Amen Amen! Isn't the Church a lovely place? Its not just for spiritual growth, she is very pragmatic too Wink

stay blessed,
habte selassie
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« Reply #23 on: May 25, 2012, 03:21:45 PM »

I realize that in a solution the pH and pOH must add up to 14 (aside from certain rare exceptions), however does the same apply to the hydroxide and hydronium concentrations in a solution? That is, do the hydronium and hydroxide concentrations have to add up to 14 or 10^-14 (I.e. ten to the negative fourteenth power) or is that not necessarily the case?

Thanks again and please pray for me,
Severian
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« Reply #24 on: June 19, 2012, 03:26:45 PM »

I got a 'B+' on my final exam for Chemistry. I am a bit upset because the only reason I didn't get an 'A' on it was because I made the silliest mistakes, which I didn't realize I had made until AFTER I took the exam. Still, it is/was a pretty hard course/exam, so a B+ isn't too bad. Thanks again for all the help the posters on this thread provided. Smiley
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« Reply #25 on: June 19, 2012, 06:03:21 PM »

Greetings in that Divine and Most Precious Name of Our Lord and Savior Jesus Christ!

I got a 'B+' on my final exam for Chemistry. I am a bit upset because the only reason I didn't get an 'A' on it was because I made the silliest mistakes, which I didn't realize I had made until AFTER I took the exam. Still, it is/was a pretty hard course/exam, so a B+ isn't too bad. Thanks again for all the help the posters on this thread provided. Smiley

That is the difference between an A and a B at university level, professionalization. If you read a published study with typos or simple errors would you trust it? If you were working on a project in the real world, would you accept your co-workers making project ending mistakes even if just simple ones? Either embrace the effort and enjoy the B, or evolve your game to Japanese levels and earn the A at every step.  Me, I think there is no shame in a B.  Admitting fault is more valuable the showboating accomplishments Smiley

stay blessed,
habte selassie
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« Reply #26 on: June 19, 2012, 08:44:40 PM »

Greetings in that Divine and Most Precious Name of Our Lord and Savior Jesus Christ!

I got a 'B+' on my final exam for Chemistry. I am a bit upset because the only reason I didn't get an 'A' on it was because I made the silliest mistakes, which I didn't realize I had made until AFTER I took the exam. Still, it is/was a pretty hard course/exam, so a B+ isn't too bad. Thanks again for all the help the posters on this thread provided. Smiley

That is the difference between an A and a B at university level, professionalization. If you read a published study with typos or simple errors would you trust it? If you were working on a project in the real world, would you accept your co-workers making project ending mistakes even if just simple ones? Either embrace the effort and enjoy the B, or evolve your game to Japanese levels and earn the A at every step.  Me, I think there is no shame in a B.  Admitting fault is more valuable the showboating accomplishments Smiley

stay blessed,
habte selassie
I agree with you to a big extent.
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« Reply #27 on: June 19, 2012, 08:48:06 PM »


Greetings in that Divine and Most Precious Name of Our Lord and Savior Jesus Christ!


I got a 'B+' on my final exam for Chemistry. I am a bit upset because the only reason I didn't get an 'A' on it was because I made the silliest mistakes, which I didn't realize I had made until AFTER I took the exam. Still, it is/was a pretty hard course/exam, so a B+ isn't too bad. Thanks again for all the help the posters on this thread provided. Smiley

That is the difference between an A and a B at university level, professionalization. If you read a published study with typos or simple errors would you trust it? If you were working on a project in the real world, would you accept your co-workers making project ending mistakes even if just simple ones? Either embrace the effort and enjoy the B, or evolve your game to Japanese levels and earn the A at every step.  Me, I think there is no shame in a B.  Admitting fault is more valuable the showboating accomplishments Smiley

stay blessed,
habte selassie
I agree with you to a big extent.

By the way, I hope my tone came across in the positive light I was aiming for.  I was trying to reiterate that an A is a big deal, and at an academic level a well earned B is perhaps even more an accomplishment worth celebrating.  As they say, when climbing a mountain, the biggest thing is not to get caught up on the hiking part, but too just enjoy the view Smiley

stay blessed,
habte selassie
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« Reply #28 on: June 19, 2012, 08:59:59 PM »

^No worries. I understood your intentions, and I agree. A 'B+' in a class of that difficulty is still pretty respectable. However, a part of me just wishes I got the 'A' that I had aimed for because I did study pretty hard. Still, I was quite satisfied when I found out my grade, and most of the grades other students received were 'B-'s and 'C's (or at least from what I could tell), so in comparison to many others, I still did pretty well.
« Last Edit: June 19, 2012, 09:06:36 PM by Severian » Logged

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« Reply #29 on: June 27, 2012, 12:13:26 AM »

For those of you who so graciously helped me on this thread, did any of you study Chemistry in college? And if so, what level of math did you need to have behind you in order to understand Chemistry? I am curious as I have taken up an interest in Chemistry, but I am not a big fan of math.
« Last Edit: June 27, 2012, 12:14:25 AM by Severian » Logged

"These things I have spoken unto you, that in me ye might have peace. In the world ye shall have tribulation: but be of good cheer; I have overcome the world." -Jesus Christ

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