How to Find Theoretical Yield (2023)
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- Опубліковано 10 гру 2024
- Convert all amounts to Moles
Divide all moles by the COEFFICIENT of balanced chemical reaction
Whichever of those results is lowest corresponds to your LIMITING reactant
Use that reactant's number of moles and a MOLE RATIO to figure out amount of product made
Convert to grams if necessary
Explained this better in 5 minutes than my professor did in an hour
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Really
An hour,the whole year!😢
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I watched 10+ tuts, and din't get any shit out of them, this explained the concept in less than 6 minutes, W professor.
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It's the day before the last day of school. I've been stuck on the last problem on my portfolio assignment for hours. I couldn't figure out theoretical yield. I watch this video and in 5 minutes and 21 seconds I finally understand. Thank you very much for this video and teaching me a problem that closely resembles the one I have to do. Thanks again!
you explained this SO clearly, i finally understand THANK YOU !!
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Understanding the yield calculation was so easy. I never even thought about it before.
I watched so many videos and could not understand how to do this until I found this one! Thank you so much!
Hey guys don't think deeply, people who work in pharmaceutical it's very easy to find.. Molecular Weight of your product/Molecular Weight of Starting Material*Batch input(Theoretical yeild), Batch input/Theoretical yeild*100= Percentage of yeild
for some reason theoretical yield is a concept i was NEVER able to wrap my head around no matter how many times it was explained to me. after literal and actual weeks, THIS is the video that's finally managed to drum it into my thick skull. Thanks man
Thanks dude! I was confused by this for years and you made it so simple.
Thanks sir for uploading such an amazing vid
You can't belive how much did your video help me... This subject has too many videos explaining too many different useless ways of solving something as simple as you just put it. Great job, you saved me!
EDIT: I have a quiestion though... what do I do if I am not given the masses of not the reactants but only of the product?
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Perfectly clear, thank you.
Magical video 📸
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W video I understand it very well thank you
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Great Explanation ❤
The explanation was great, I finally understand it. Thank You. BUUUT, where did you get the initial 3.0g for 2H2 and the 29.0g of O2. Can you please explain? Thank You,
It's part of the question
Great explanation this is the only video that’s helpful for me
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Best xplantion of all time for theoretical yield I also want actual yield how to find it
Really great explanation!! Thanks so much❤
very well explained!!
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what if you not given the mases of the reactants only one?
Then your screwed
Then the reactant about which information is given is your limiting reagent
very useful! Thanks a lot
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I appreciate how you have presented and simplified this concept ❤....but I a little bit don't understand on the part where you are dividing number of moles by their respective coefficients😮
love it
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Good video, bit wanted to ask that what if we don't know the mass of one reactant
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Damn that was faster than the other formula and much easier
thank you!
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Among the reaction which reactant are we suppose to use for us to find theoretical yield
Isn’t the molar mass 2H2 4?
Or are we ignoring the mole?
How would I find the percent yield with this kind of equation, where neither PY or actual Yield is given. Would the AY be molar mass of a product - the amount of product lost?
But why dindnt you multiplythe 2 of H2O by the 1.5 of h2
Yes! Wonder why this wasn't the case?
@chemistNate what happens when you three products
Wouldn't o2 be limiting cause it's completely used up
I tried doing this with the net ionic eqn but it didn't give the same mass of product in the end. Strange
It's for a quiz retake im doing with the reaction between 0.040 moles of Silver I Nitrate and 50.0 ml of 0.60 M Calcium Chloride.
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Couldn't understand better
How did you find 3.0 grams from hydrogen
it was given in the question
where the hell did you get 18.0 grams from?
The molar mess of H2O
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