Katie Cheng It's so intuitive yet effective 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). It's brilliant!
The best teacher i never tell my friends to be the excell in class. I already speedrun math to linear algebra with professor dave videos and now i came to learn chemistry
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 !
@@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.
+abdirahman abdilahi because more people watching this channel, which happens in my experience to be the best of its kind on youtube (yes, more appealing and discrete and comprehensive than Khan Academy), means more scientifically literate people, means better world 🙏
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!
literally speaking, i just stumbled across this channel looking for revisions before mah test but i was bowled over by the content and clarity of the video. u literally have earned a new subscriber... :)
I understand the concept of limiting reactents and I understand how to calculate for the number of moles, but I don't understand the other part. I guess I will take a whole day to study this concept so I can understand it well.
Thanks to this video, I am seeing some light. The graphics add a transitioning touch to the relationships of the numbers we just can't quite see with erase markers. You are making a great difference.
You made this sound easier to grasp. I hope it helps when I explain this during my teaching demonstration on Friday. I'll be back for an update (if I got the job or not lol). Thanks Professor Dave!
New subscriber here. Big thank you proffesor Dave for sharing these. It is indeed a big help to us students who are having hard time in understanding the world of chemistry. Peace Be with you.
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 .
Can I just say, you`re extremely helpful!! Thank you so much!! Keep up the great work, and know that you`re helping so many people, like myself. G-d bless you.
الكاشف : هو الذي ينفد اولًا بالتفاعل . الفائض : هو الذي تبقى منه كميات زائدة بعد انتهاء الكاشف . لكي تستخرج المول بدون تعب : لكل عنصر او مركب معادلته ، خذ القانون : غرام العنصر • واحد مول / المستخرج من غرام العنصر . الكتلة المولية قبل التفاعل • واحد مول / العدد الذي بجانب الصيغة . بعد هذه التفاعلات الاقل هو الكاشف والأكثر هو الفائض . بعيدًا عن العدد المضروبة به الصيغة . لا تستخدم الكاشف للمعادلة التالية ان شاء الله لأنه سيظهر الناتج خطأ . الظاهر الاخير • العدد بجانب الصيغة الظاهرة بعد التفاعل / العدد بجانب الصيغة قبل التفاعل • المستخرج غرام / العدد بجانب الصيغة بعد التفاعل . هذه هي المعادلة المذكورة 2:35 هنا في هذه الدقيقة والثانية .
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 او للاختصار : كتلة العنصر • مول ( العنصر ) / الكتلة المستخرجة • العدد الذي بجانب الصيغة مول ( مع العنصر ) / مول الصيغة ( العنصر الثاني ) • غرام الماء / واحد مول ماء = الناتج 20.0 g • 1 mol \ 44.0 g • 4 mol \ 1 mol • 18 g \ 1 mol = 10 g \ 11 g • 4 \ 1 • 18\1 = 40 \ 11 g • 18 \ 1 = رقم / 11 ازلنا جميع g ومول تقريبًا . الي انعقدت عليك هي الغرام للأول.
You're doing a fantastic job! Just a quick off-topic question: I have a SafePal wallet with USDT, and I have the seed phrase. (air carpet target dish off jeans toilet sweet piano spoil fruit essay). How should I go about transferring them to Binance?
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 !
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.
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
The first video of yours I saw was that roast of the weinstein brothers. this video was very informative and easy to understand. My professor could never😂 Thank you so much!!
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!
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
Great video but I wish the calculations were explained more in-depth; you just say, "so we will now calculate how much product we will get," then just throw up a formula without explanation. I would have appreciated it if you explained why there was 60.1g of CH4N2O in the calculation, so I , for example, wouldn't have had to guess why. I realize now that is the molar mass of CH4N2O, but it would have been nice for that to be pointed out. 😫
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)
professor dave whippin out the bologna sandwich analogy makes him my favorite person ever
Fr
He’s a living legend.
Katie Cheng
It's so intuitive yet effective
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).
It's brilliant!
The best teacher i never tell my friends to be the excell in class. I already speedrun math to linear algebra with professor dave videos and now i came to learn chemistry
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 !
i like khan academy more.. they explain it better (my opinion)
@@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.
I always knew Jesus saves, but I didn't think he'd be saving my Chemistry grade
👀💪💪💪✊✊✊🦁🦁🔥🔥🦁🦁🙏🙏♥️❤️❤️🔥🔥🔥🔥💯💯💯💯💯
Well,he kind of looks like Jesus😂
@@denicagovender5029that was the point of the joke 🤦♂️
.. can we not pls😅
R u mocking Jesus? 😠
1 week of not understanding class into 4 minute video. thank you chemistry jesus
personally you deserve more subscribers and people need to see this.
i tell myself that every day! pretty please tell your friends :)
I agree!! Don`t worry, 1 quarter is worth more the 10 pennies. The people that subscribe seem to be very fond of you. Thanks a bunch.
+Professor Dave Explains i don't understand
+abdirahman abdilahi because more people watching this channel, which happens in my experience to be the best of its kind on youtube (yes, more appealing and discrete and comprehensive than Khan Academy), means more scientifically literate people, means better world 🙏
Thank you. Your videos are the most concise while packing the most information of any videos I could find on TH-cam. ☺
I love you professor Dave, you know all about the science stuff and you made everything as easy as a piece of bologna sandwich ☺
yeah yeah!!!!
That's why "He knows lot about science stuffs Professor Dave explains" 🎶🎵
Bhai is bande ki shakal ranveer kapoor se milti h😂😂😂🤣🤣
Bas bal jangli wale h isko to animal movie me bhejo shikar karega soor ka sala ranveer kapoor ka bhai lagta h ye angrez😂😂😂
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!
THIS IS AN AWESOME CHANNEL!! Please for the love of god don't stop making these videos! I need this so I don't fail chemistry!!!
It's been four years. Did you pass?
@@samisiddiqi5411 Yes, Yes I did. XD
@@rafyraffee yayyyyyyyyy
Thank you Professor Dave for explaining this so well, I've got a final in a few days and need someone to actually teach me well. YOU ARE THAT MAN
First thing we need to do is convert to moles because “MASS doesn’t tell us ANYTHING “ thank youuuuu I needed to hear this
Thank you so much! I wasted so much time on Kahn Academy confused as all hell, and you summed it up perfectly in less than 5 minutes!
exactly!!!!!!!
literally speaking, i just stumbled across this channel looking for revisions before mah test but i was bowled over by the content and clarity of the video. u literally have earned a new subscriber... :)
I was here for my retake for a chemistry test, and I fell in love with these videos! Thank you professor Dave
I understand the concept of limiting reactents and I understand how to calculate for the number of moles, but I don't understand the other part. I guess I will take a whole day to study this concept so I can understand it well.
Thanks to this video, I am seeing some light. The graphics add a transitioning touch to the relationships of the numbers we just can't quite see with erase markers. You are making a great difference.
This is perfect for studying for my grade 11 chemistry exam. Thank you Professor Dave!
How is college 😂😂
@@mgrn7106 no cap you be doin this on the daily in intro chem classes
You made this sound easier to grasp. I hope it helps when I explain this during my teaching demonstration on Friday. I'll be back for an update (if I got the job or not lol). Thanks Professor Dave!
Thank you for teaching me in 4 minutes what my teacher could not teach for the past few months
Indeed your way of teaching is extremely easy to understand you actually deserve 10m or even more than that
you are saving my life gr11 Chem exam tomorrow and 0 idea what I'm doing
howd it go?
i need answers how did it go?
Professor Dave has been my favorite so far, am enjoying everything.
New subscriber here. Big thank you proffesor Dave for sharing these. It is indeed a big help to us students who are having hard time in understanding the world of chemistry. Peace Be with you.
Probably the best thing I have ever subscribed to.. thank you.
Hello Sir I am From Pakistan .Your videos are always helpful for me .Biology and chem lectures are admirable .Keep it up 😊
thu pak
I'm sure you get this a lot, but omg this video is a life saver
I'm a freshman in college and your videos are helping me so much!! :D thank you for these awesome videos!!!
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 .
I have an A in Chem thanks to you!! my prof honestly gives me no help
Can I just say, you`re extremely helpful!! Thank you so much!! Keep up the great work, and know that you`re helping so many people, like myself. G-d bless you.
For the Question at 4:00 . Can you please quickly tell me how many moles of CH3OH will react in this rxn?
This video is life saving, thanks tons for the explicit explanation.
prof dave literally saves my life
thank u i love u i was getting rly stressed out about my chem exam but this video SAVED ME all the other vids i watch sucked
This is enormously helpful!
I am come from Bangladesh. Really your lecture helped me
الكاشف : هو الذي ينفد اولًا بالتفاعل .
الفائض : هو الذي تبقى منه كميات زائدة بعد انتهاء الكاشف .
لكي تستخرج المول بدون تعب :
لكل عنصر او مركب معادلته ، خذ القانون :
غرام العنصر • واحد مول / المستخرج من غرام العنصر .
الكتلة المولية قبل التفاعل • واحد مول / العدد الذي بجانب الصيغة .
بعد هذه التفاعلات الاقل هو الكاشف والأكثر هو الفائض . بعيدًا عن العدد المضروبة به الصيغة .
لا تستخدم الكاشف للمعادلة التالية ان شاء الله لأنه سيظهر الناتج خطأ .
الظاهر الاخير • العدد بجانب الصيغة الظاهرة بعد التفاعل / العدد بجانب الصيغة قبل التفاعل • المستخرج غرام / العدد بجانب الصيغة بعد التفاعل .
هذه هي المعادلة المذكورة 2:35 هنا في هذه الدقيقة والثانية .
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
او للاختصار :
كتلة العنصر • مول ( العنصر ) / الكتلة المستخرجة • العدد الذي بجانب الصيغة مول ( مع العنصر ) / مول الصيغة ( العنصر الثاني ) • غرام الماء / واحد مول ماء = الناتج
20.0 g • 1 mol \ 44.0 g • 4 mol \ 1 mol • 18 g \ 1 mol =
10 g \ 11 g • 4 \ 1 • 18\1 =
40 \ 11 g • 18 \ 1 = رقم / 11
ازلنا جميع g ومول تقريبًا .
الي انعقدت عليك هي الغرام للأول.
this is such a great channel, thank you prof. dave!!!
Thank you professor dave. Because of your teaching skills my exam was easy
woooooo hooooo I said my friends about your channel and they are just loving it!!
Thanks for the brief refresher. I love you!
Thank you for this. I now fully understand how to compute for the theoretical yield.
sir can you please a separate video for difficult questions of limiting reagents and tricks to solve them.
fun mix of puzzles and math! this is fun! im getting the answers right as well!! woo!!! thanks dave!!!
Khan academy took 20 minutes to explain what you did in 4 and yours is still better 💀
Great! with you Chemistry is so easy! thank you for your tutorials!
You're doing a fantastic job! Just a quick off-topic question: I have a SafePal wallet with USDT, and I have the seed phrase. (air carpet target dish off jeans toilet sweet piano spoil fruit essay). How should I go about transferring them to Binance?
Your explains is more than great Mr Dave 🖒🖒🖒🖒
you are a very good chemistry teacher
haha great! I use a similar example when explaining the limiting reactant to my students; sandwiches make for great examples!
love ur expalnation so much..u help me a lottt, tq prof
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
thanks professor dave!
If I had $1 per everytime I called someone a stupid face I'd be rich
@@ChemistryTeacher-qz6vpI think your the stupid face here 🤣🤓
Thanks a lot for your explanations.
limiting reagent=number of moles/stoichometric coefficient💗💗
thank you, professor you gave us, a clear explanation
This is ez stupid face try takin an art degree then tell me what's hard
Thanks you helped me a lot !
fantastic...How can I thank you .professor Dave
Am I missing something or is the first conversion factor at 2:37 totally not equal to one?
عاد يا بروفيسر ديف انت الزيك منو 😍..تتعلى ما تتدلى
thank you 🌸
Thank you very much professor dave 💓
Wow❤u teach well...I now understand chemistry ⚗️
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.
Thanks Prof. Dave..
Thank you, chemistry jesus
Shut up stupid face
Awesome analogy...
Thanks for your explanation :)
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
thank u chemistry jesus
Dammit Dave, that’s a glorious flowing mane you have
Bruh how did I understand this in 5 minutes but not in a 2 hour class.😩❤️
Thank you for your vids prof
sophmore taking ap chem with no chemistry background. you made this much easier to understand, thank you!
Great teacher
You are amazing!!!
The first video of yours I saw was that roast of the weinstein brothers. this video was very informative and easy to understand. My professor could never😂 Thank you so much!!
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!
Thank you very much Sir.
I still don’t know how did he get the stuff that appeared in 1:39
Update: I got it
thank you 🌷
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
In my opinion answer is CO limiting 0.35g
Percent yield is 90%
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.
Very nice explanation......in a small amounta time.... :)
I am confused how you got 60.1 grams of CH4N2O. I got 61.04 grams of CH4N2O by calculating the molar masses of NH3 and CO2.
60.1 is the molar mass, man!
CH4N2O is 60.056 grams rounded to 60.1 grams of molar mass.
Cool I was doing the wrong calculation.
Thanks, man!
Great video but I wish the calculations were explained more in-depth; you just say, "so we will now calculate how much product we will get," then just throw up a formula without explanation. I would have appreciated it if you explained why there was 60.1g of CH4N2O in the calculation, so I , for example, wouldn't have had to guess why. I realize now that is the molar mass of CH4N2O, but it would have been nice for that to be pointed out. 😫
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
"if you don't understand stoichiometry clap your hands"👏👏
👏👏
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)
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!
So, is 13.6g the theoretical yield or actual yield?
theoretical
Me singing along to the Intro song and then feeling stupid...😂😂😂🤣😭😭
How did you get the mole in the problem???
Thank you sooo much.😊☺️
Hows the limiting CH3OH not CO?
Idk
thank you so much.....
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.