STOICHIOMETRY - Percent Yield Stoichiometry Problems - CLEAR & EASY
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- เผยแพร่เมื่อ 23 ส.ค. 2024
- STOICHIOMETRY - Percent Yield Stoichiometry Problems - Short video explaining how to determine the percent yield in stoichiometry problems involving mole ratios! Stoichiometry can seem difficult, but I'll show you how to solve problems easily! Check it out.
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That's a huge compliment! Thanks You. I'm so stoked I was able to help!
Wasn't it 300 g not 301 ??
And 251 g of H20 instead of 250...
@ 7:02 where did you get 338 from?
Since 301g/32g X 18g = 169g
So clearly explained! Thank you
Haha, you're probably right. Good job paying attention. Thanks for the compliment bro. If the video was really helpful, then please share it, pass it along. Tune back in when you some help again. Peace.
Thank you but can you explain again how to get the 297 plz
This video was really great. I didn't understand the personalized video my teacher sent me.
Why is the limiting reactant the H2???
Probably you already found out the answer to your question by now, but for the sake of those people who'll be having the same problem as yours in the future, I'll explain. :)
As you can see in the chem eq, we have 2 mol of water so your solution should be, 301g O2/32g O2*2mol H2O*18g H2O
Hope this helps :)
thank you so much for the amazing video, but do we have to put the percentage in the proper # of sig figs ? please reply. thank you !
No
At 7:00 into the video I solve for the limiting reactant. Hydrogen produces less H2O at this point, than O2 can, so H2 is called the LR. O2 has the potential to make more h2o so it is the excess reactant. I hope this helps, perhaps watch the video again or ask me again, and cite a time in the video where you get confused. Best wishes.
I am a little bit confused, I calculate the mole of H2 and O2 and it appear that O2 is limiting... I also check with other people and they've all said O2 is limiting. So now I'm stuck with which is limiting and which is excess :(
This was really helpful, thank you for posting this video! :)
Thank you so much
ive noticed that the oxygen in grams changed from 300 g in the problem to 301g. I dont know if it really matters but just wanted to let you know. Otherwise great problem! I have my second exam next wednesday!!!
Great explanation! Few mistakes but regardless it really helped me out! Thanks so much! :)
Thx for the compliment!
why did you use 1 mol of H20 as a conversion factor when it was 2 mols H20 in the equation?
I was wondering the same
this is what I was looking for. thank you.
this helps me so much, thank you!!!!
how would I go about solving a percent yield 28.1g of Sb4O6 and excess C if 17.3g of Sb is recovered along with an unknown amount of CO
thank you!!!!
alright dudes lets bust this out
Great video!
Very helpful!
Right on!
Thank you!!!
You said have a good night but I watched it in the morning. Is it still good night
All good. Good morning to ya!
Going backwards screwed me up
bust out that calculator XD
Up the Volume, this vid feels quiter than the others
Too soft spoken volume at 100% and I am still straining to hear you.
Sorry about that. Budget microphone :( Later videos have better quality.
Ronna Blackley ,do you know McConahein formula,calculate Av Al2O3 from Al2O3?