Limiting Reagents and Percent Yield
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- เผยแพร่เมื่อ 28 มิ.ย. 2024
- Chemistry doesn't always work perfectly, silly. Molecules are left over when one thing runs out! Also we never get all of the products that we thought we might by doing the math. You gotta know about the limiting reagents and the percent yield! Don't worry, it's as easy as bologna sandwiches.
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Some sandwiches are just Bologna(O) and bread(H), others are double or triple decker and have a different medium like... peanut butter (Ammonia), or have different things that act as bologna but aren't or dishes that don't need it (a turkey sandwich or literally just table salt).
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Came here from Khan Academy. Much clearer presentation than the current Khan Academy video on limiting reagents. Thank you so much!
I actually didn't understand this video, so I went to the khan academy video after reading your comment.. to see if it would be any different!
Strangely enough, I understood their explanation! KA video is more detailed than this one is, I don't understand why you guys didn't like it!
In any case ,I'm greatful to both these educators !
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@@adarsh65kumar Perhaps learning the same thing from multiple teachers helps better, just a hypothesis though, it happens to me too. If I learn something from one video and don't get it clearly enough, I watch a different one on the same topic and it gives me a different perspectives on the same topic, making it a bit easier to understand.
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Limiting Reagents and Percent Yield are one of the most confusing topics for me in Chemistry. Thanks to you, I understand it better. This is really helpful for students like me. Thank you very much!
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Thank you for this. I now fully understand how to compute for the theoretical yield.
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sir can you please a separate video for difficult questions of limiting reagents and tricks to solve them.
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haha great! I use a similar example when explaining the limiting reactant to my students; sandwiches make for great examples!
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thank you so much.....
How did you get the mole in the problem???
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so for the same reaction, if i change the masses of the reagents, could the limiting reagent change? or is it always the same for the same reaction?
yep it totally depends on how much you have of them!
quick question, when we getting the theoretical yield for CH3COOH, why we taking 21.4g as our final answer, not 21.42g. I went back and rewatch the sig figs chapter, but it still couldn't help me with the result. Thank you in advance for helping !
its 3 sig figs so you wouldn't add the 2
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someone please help why didn't we multiply ammonia with 2/17? because according to the equation its 2 moles of ammonia reacting right?
Do you HAVE to find out how much of both reactants can react with each other to subtract and find the excess.
Well like the amount of excess
Very nice explanation......in a small amounta time.... :)
Sir if u have time please try to make a video on equivalent concept in chemistry really hard to get feel and intiution of that topic I would be very very thank ful to you ( from India Telangana)
Wow❤u teach well...I now understand chemistry ⚗️
Am I missing something or is the first conversion factor at 2:37 totally not equal to one?
why did you round two sig fig, when calculating?
no reason, just to keep it tidy
Dammit Dave, that’s a glorious flowing mane you have
Where were you the whole year good sir...
At, 3:59 is the chemical CO , Carbon and Oxygen or Cobalt?
CO is carbon monoxide! All elemental symbols have just one capital letter, so cobalt is Co.
Khan academy took 20 minutes to explain what you did in 4 and yours is still better 💀
how did you get the 0.454 moles of NH3?
that's just seeing how much NH3 would react with the CO2 given the stoichiometric ratio
Hi, professor Dave. May I ask you why the answer of the question is 89.3, not 90? I thought sig fig would be 1 since the one with the fewest sig fig is 100 (1 sig fig). Sorry if the question is too silly :( and I very apprieciate your well-made videos.
if anyone else could reply to me, I would also very apprieciate.
holy shit bro ty
Bruh how did I understand this in 5 minutes but not in a 2 hour class.😩❤️
Guys pls someone explain how to find the actual yeild if not given
The final percent yield 89.3% is a little off because of a rounding error: 19.1 ÷ 21.42857 ≈ 89.1333% ≈ 89.2%. Don't round to sig figs too early in the process and you should be fine 😊
when converting to moles in the ex. in 1:45, why can't I convert between the substances after I have just the first substance in moles? and also why do I get different result while doing it in comparison to just calculating it separately?
10g NH3* 1 mole NH3/ 17g NH3= 0.588 moles NH3
0.588 moles NH3* 1 mole CO2/ 2 moles NH3= 0.294 moles CO2
--------while calculating separately will give me 0.227:
10g CO2* 1 mole CO2/ 44g CO2= 0.227 moles CO2
ok so the answer is because if I do that, I will get the amount of the other substance in moles that will theoretically react flawlessly with the first substance and not the actual amount of the substance.
I don't know if anybody needs it but there you have it
Legend...
Hows the limiting CH3OH not CO?
Idk
Molecular mass of ch4n2o is 16+24+16= 54 g/ molecular but how did 60.1g is multiplied with .227 moles ?? Please explain how did 60.1 comes
it's the molecular mass, you miscalculated it.
Sir please please I beg your pardon infinite please video on equivalent concept in chemistry
رقم افوغادرو 6.022×10²³
رقم ٦ هذا المقصود به كاربون .
الكتلة المولية : العنصر × كتلته .
C⁶ × 12 .
عندما يحصل تفاعل كيميائي العنصر يتغير من جهة اليسار إلى اليمين اي النواتج ، لذلك يجب حساب مول جديد وهذه هي الطريقة :
- قبل التفاعل الكيميائي
البروبان × عدد المول / الكتلة المولية = المول .
- بعد التفاعل الكيميائي
المول × عدد المول الجديد / عدد المول القديم
المول المحصول × عدد البروبان / عدد المول .
عاد يا بروفيسر ديف انت الزيك منو 😍..تتعلى ما تتدلى
if during the experiment, you only included 97% of grams of 1 of the reactants, would you still get the same substance you’re expecting? only that, you wont get the theoretical yield amount in grams?
OR the whole experiment would be wrong because the expected substance wont happen?
oh wait!!! i think, we can still get the same substance for as long as we have the reactants necessary. the change in grams in the reactants does not affect the expected substance, only the amount in weight afterwards. please tell me if im right or wrong, Prof Dave. :) if im right or wrong, why or why not
@@adr5617 That’s right. Change the amount of inputs only affects how much of the product you get. In the bologna example, no matter how much you change the amount of bologna or bread, you always end up with bologna sandwiches.
Nicely done, you did not show which was the limiting reactant but it's CO
i put a box around it!
but both ch3oh & co equate to 0.357 mol. im confused why you chose co as the LR. need clarification on this one prof. dave.
thanks in adv.
sophmore taking ap chem with no chemistry background. you made this much easier to understand, thank you!
Professor Dave is the best of the best... oooops
I still don’t know how did he get the stuff that appeared in 1:39
Update: I got it
So, is 13.6g the theoretical yield or actual yield?
theoretical
Well i don't know how much he really knows since he reads off a prompter
How to get the 0.294 mol CO2
0.588 mol NH3• 1mol CO2/ 2mol NH3
This is great work, but unfortunatley it would be nice if the problem showed how the understanding of dimensional analysis played a role. But other than that it helped me get to an answer!
i have a separate clip on dimensional analysis, take a look!
The molar mass came out to be 64 g/mole for CH3COOH, what did I do?
I dunno! You must've counted wrong. You can email me your calculation if you want.
Whoops, I see