Limiting Reagent and Percent Yield

[@slightlyferalraccoon]

Thank you so much for actually going into detail and explaining everything thoroughly. I am a fast learner but I have to understand every aspect of a problem at all viewpoints.. Because of this, it's frustrating and I often have to teach myself. I think you will be my new chemistry teacher for the rest of the school year from this point forward. :) I cannot thank you enough! Wish me luck on my exam tomorrow, heh. :)

[@BensChemVideos]

Good luck!

[@Multi1Vids]

I cant believe how much i learned in 12 minutes as appose to a month...

[@blablabla12373]

This video was so helpful, my teacher didn't really do a good job explaining how to do this in class and now I completely understand and know how to do it. Thanks so much!

[@Sarah-kd7qg]

By far the best video I’ve stumbled upon clearly going through the limiting reactant question! Thank u soooo much

[@devorahtabak3704]

Clearest video ever! Ben you are the clearest teacher! Please become a professor somewhere soon so you some students out there could get real lucky!

[@DeniseDarell]

ive learned more in 12 minutes than i have for the past month and a half. praying i do good on finals. thanks.

[@naimaA23]

Thanks a lot! One of the few people who can teach this concept quite well. Well done!

[@deadchannel772]

I'm going to learn this in my next honors chemistry class (I looked ahead to see if I can get a better understanding before I learn it). Thank you very much. Now I know what I'm doing before the other kids in my class. You're a magnificent teacher :)

[@askfritzaboutmaths3874]

great explanation, dont be nervous because you must understand it very well to explain it so simply.

[@discodancer22]

learned more here than hours spent in class. thanks man

[@Fizzle387]

First video I found that isn't filled with unnecessary fluff or too fast talking, kind of understand now ,thanks

[@sarabreshears6767]

OMG three months of studying this stuff and I don't get it. 12 minutes and It makes so much more sense, thank you!

[@katelynprice9529]

Yet again, you saved me from failing. Thank you so much. I think I will be using a lot more of your videos throughout this year. :)

[@musamustafa3390]

this is the best explanation of this section of chem. i have ever seen! god! finally!

[@sarahmcglumphy8889]

Bless your soul for helping me pass Chem 211, would rather have you as a teacher than my current professor...

[@rambone81]

I rather appreciate your videos on chemistry, they have helped me out considerably. I ended up with a professor that I have a hard time understanding the accent of which makes it difficult to know what is being said in the lectures. Your videos have helped to clear up my confusions. I would recommend though a title change to be inline with the text book of "intro to chemistry - limiting reactant and yield calculations". This may help others find your video easier so you can be of help to more.

[@ShauryaMago]

The concept is more important than the actual value.

[@jravila65]

Youre like the PatrickJMT for chemistry! Thank you very much youve just replaced 3 weeks of missed lectures in 12 mins!

[@sharojany534]

You are such a great person........thanks alot ...please give lecture to our lecturers. They should learn how to teach.......BIG THANKS..........

[@CaptianComment]

We literally spent a whole week doing what took him 12 minutes to explain. Bravo, old sport.

[@jacklavelle1716]

i spent so much time trying to understand this because of you its easy

[@lisamcarthur6761]

you're amazing. I need someone smart like you in my life to make sense of all this nonsense. thank you.

[@3011remaerdyad]

KZbin teaching betterthan my teachers for forever

[@theomarman1]

Thank you I was going to lab when I watched this, you are a life saver!!

[@saraht.8626]

Thank you so much for yr videos! I love how clear/direct you are with all the steps that you take in a chemistry problem. It really helps me to break it down in my head and to fully understand what is going on. Chemistry is really hard condensed in 6 weeks, but you make it easier for me! Thanks again c:

[@trinklest]

Thank you! I thought I understood it in class, but I just couldn't figure it out at home. I finally am able to visualize the problems now! 😊

[@tolessaegeta73]

from your teaching I have learned some steps please continue.

[@NoobityBoobity]

Saving my life once again. Round 2 with O-chem 202 and I ain't losing

[@travisfabisch9383]

have a big chem test coming up and this really helped!!! thanks for posting!

[@dj.frankieg]

Thank you for the video, you will for sure help me tomorrow morning while I am taking my final.

[@MrDay515]

Thanks man, a big help. You're doing good things.

[@frodriguez9702]

my teacher just said use this, you taught me so much

[@matucky81]

way better than my chemestry teacher!! thanks alot!!!!!

[@winterwraith8339]

from what I've learned in life is that if you love something you should do your best to care for it. for example you want more ducks to get to a pound because you love seeing them so happy there and the ducks want to b in the pound too but there is a object blocking most of them. so if I help them out and eliminate that object they all can get to the pound and we will all be happy. how I feel about this guy, I love your videos and I want you to continue them but to do so you'll need help because you can't live on videos with no food or life(the blocking object). so I'll offer you some cash not much but some. it's the least I can do for all the help you've given me. Ty!!

[@BensChemVideos]

Ghost Angel thank you for your generosity and being willing to put your money where your mouth is to show your gratitude. Within a week or two I'll be uploading a video showing how to make a donation to support this channel. Thanks again, -Ben

[@winterwraith8339]

Ben's Chem Videos No problem dude. keep up the good work!

[@misterman3791]

You teach better than my Professor. Thank You

[@aliciasvoboda1450]

try learning this shit as s junior in highschool omg

[@Gam3r5t93]

+Alicia Svoboda exactly Im a junior too and i can never understand this in class, i always have to study myself to really understand

[@arjunchopra762]

god bless whoever made this. I might do good on my chem test!

[@jpmex16]

Thank you so much for your help. I know this video is old but, it still helps!

[@kaustshroff381]

I had to watch it twice, but awesome video!!!! Thanks a lot man.

[@chfzfdhh878]

thank you. thank you. thank you. Oh ma Lord I missed three days of school right before the test and missed this whole subject so again thank you

[@ScarOfHell]

thank you so much...you have no idea how much this helps :)

[@ccr5-328]

propane and propane accessories

[@chrisonliine]

Don't you mean pro-PAIN

[@dylanspinas7358]

Hey man, thanks for the help- make it sound 10 times simpler than my Chem teacher does!

[@icestripe]

Bravo, sir! Thank you for the brilliant explanation!

[@louiseheller4602]

Thank you so much Ben. Really helpful.

[@Revoluus]

Thank you so much - you cleared up a gap I had.

[@samclubb4076]

This video was really helpful! Thanks!

[@madelynmorgan2999]

thank you so much dude!! I understood this in class but was confused when studying for the final

[@ch3rbet]

Thanks for the video! I kept on getting confused because I thought that the limiting reagent was the smallest one, not the one that runs out first...if that makes sense.

[@teabrickss]

i have finals tmrw and this sort of helped. thank you.

[@JuanyJuanes]

I am so glad I never have to do these again. Thanks for helping me last semester tho

[@nplakun]

Ben. You are the man. Thanks for your videos.

[@DayanaChavez13]

I was absent from school for a week and i have a test tomorrow on this, I understand it though.Thank you so much sir

[@ricardofarinas7674]

thank you so much man, great explaining!!

[@taliebert]

OH MY GOD IT MAKES SENSE NOW! THANK YOU!!!

[@ofentsetshepe]

do we ignore the co-efficient of oxygen 5 when we calculate the molecular weight of oxygen ?

[@Sushiii911]

Thank you for uploading! Very helpful!

[@Leila_572]

Thank you very much! It was very helpful!

[@jessicahamilton4970]

thank you!!!! I have a test tomorrow and this helps!

[@ThatDirtyHippy]

Thanks a lot man! Really helped me understand!

[@Gam3r5t93]

This video saved my grade in AP Chem

[@guitarttimman]

Actually, if you did the first part of the C3H8 + 5O2 = 3CO2 +H2O and you started with the Oxygen, then you calculate that you will get .05 moles of C3H8, but you have 0.25 moles of C3H8 available, therefore Propane is in excess and oxygen is the limiting reagent. Yes that is correct. O2 is the limiting reagent.

[@MrYazid30]

THANK YOU SO MUCH.... helpful...

[@hannahbruno1682]

Very helpful! Thanks so much! Chemistry homework = done

[@ballalakers24]

wow thanks alot, extremely helpful.....keep up the good work

[@srojorules]

Very helpful indeed, thank you!

[@chrisschmid111]

I Love you this was so amazingly helpful thank you!!!!!!!!

[@ijenisecr]

Thank you so much this video helped me a lot!!!!

[@Emma-kf1ri]

SO HELPFUL. THANKS SO MUCH

[@MrsHaZohariym]

How would you do the limiting reagent if you are working with more than 2 compounds? Would you just add in the ratio of the second compound reacting with the reagent being tested?

[@98Impossible]

Thank you! Very helpful!

[@jessicaescober9894]

This is really clear! Thanks :)

[@MrRightNow96]

you just found yourself a new subscriber my friend :)

[@anniereilly7897]

Thank you so much. Very helpful

[@maureenseverance2102]

Wow this guy knows how to explain it! =D

[@3frodoysam]

fun stuff!!!

[@tas64]

another way to do this if your confused is to set up ratios and cross multiply. to find the mass of h20 make a mole ratio. (4/5=x/1) and solve for x which is the moles of water. then convert into mass.

[@greyhoundgirl103]

At the end why do you use 1 mole of Oxygen instead of 1.25 moles of Oxygen? Were did the 1 mole come from?

[@samahsamiee9659]

i think i finally understand thanks to u

[@tyeecehensley4282]

I used the mass ratios instead of mole ratios for these calculations at my school and they worked out to be the same 44g/mol of C3H8 ---> 160g/mol of O2 11g of C3H8 ----> X g of O2 X = (11x160)/44 X=40g of O2 Therefore oxygen is the limiting reagent as we need 40g of it to react completely with the propane and we only have 32g

[@naruhinafan100]

yea thanks I like this way more better

[@jacqueline8553]

very helpful video! thank you!

[@TheVDROD]

He went the step further to show that just because the amount of moles is less, it doesn't necessarily mean that it is the limiting reagent. Like he said with 0.25 moles of C3H8, you would need 1.25 moles of O2 in order to use the entire amount of propane.

[@namjin8281]

Thanks this was a great help:)

[@askachub88]

Super helpful thanks so much

[@neurokid1]

Thank you for this!

[@downrightawesome123]

Dude u rock i totally get this thanks for the help

[@rongfeng8368]

thank you so much for explaining this.!! really appreciate it. +1 LIKE!~~

[@Lola-vf7km]

You lifesaver, you.

[@jharding578]

Video helped out a lot!!!

[@vikrampinto]

very clear...thank you

[@shaneschmale]

I am the slow one in the class... this really helped.

[@Jfarmerthunder]

very helpful ! thanks

[@evanemerson9047]

why would you use O2 in the first example instead of H2O? Is their any reason or is ust the one you picked

[@dawnfmEnthusiast]

Hey man, nice video. Ben's Chem Videos Can you quickly explain how you actually record your vids - like how do you place the camera and stuff. I would be very grateful for that. Thanks.

[@ShreyaRawatdocshreerawat]

Really it helped a lot...

[@KateDensley]

so helpful thank you! Finally understand! :D

[@amaskylom]

You saved my life.

[@irena5257]

thanks for the clarification!

[@bellababy812]

Great video thank you so much!!!

[@JZXKenny]

So just to be clear, we can't actually calculate actual yield unless we're finding it through a real experiment, right? If I had to find percent yield on an exam, I can assume the teacher will provide me the actual yield, right? Sorry, sounds like a stupid question but this is all new to me

[@Justmetmt24]

Yes. There's no way to calculate the actual yield without physically doing the experiment, so it should ALWAYS be given to you.... in an exam.

[@DunWannaTellMyName]

thnx bruv :)) rly helped n_n especially the PB and jelly example :P

[@Intracranial110]

Hey friend can you explain why 1.25 mol O2 would be the LR. I would have pick O2 based on 1 mol O2. I dont understand this, I need a number to compare. Thanks for your help

[@berkcanozkara7316]

ty so much it is really helpful