That's a huge compliment! Thanks You. I'm so stoked I was able to help!

So clearly explained! Thank you

this helps me so much, thank you!!!!

Thx for the compliment!

this is what I was looking for. thank you.

This was really helpful, thank you for posting this video! :)

Great explanation! Few mistakes but regardless it really helped me out! Thanks so much! :)

Very helpful!

Great video!

Thank you!!!

Right on!

Thank you so much

This video was really great. I didn't understand the personalized video my teacher sent me.

thank you!!!!

@ 7:02 where did you get 338 from? Since 301g/32g X 18g = 169g

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 !

@2oqh

2 жыл бұрын

No

Why is the limiting reactant the H2???

Wasn't it 300 g not 301 ??

@blackestbear6631

6 жыл бұрын

And 251 g of H20 instead of 250...

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.

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.

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!!!

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

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 :(

alright dudes lets bust this out

why did you use 1 mol of H20 as a conversion factor when it was 2 mols H20 in the equation?

@jungy9001

8 жыл бұрын

I was wondering the same

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 :)

You said have a good night but I watched it in the morning. Is it still good night

@sciencepost

6 жыл бұрын

All good. Good morning to ya!

Up the Volume, this vid feels quiter than the others

bust out that calculator XD

Going backwards screwed me up

Too soft spoken volume at 100% and I am still straining to hear you.

@sciencepost

10 жыл бұрын

Sorry about that. Budget microphone :( Later videos have better quality.

@vanmenggolo58

7 жыл бұрын

Ronna Blackley ,do you know McConahein formula,calculate Av Al2O3 from Al2O3?