September 30, 2024, 11:31:02 AM
Forum Rules: Read This Before Posting


Topic: Help on problems dealing with Percent Yields  (Read 3612 times)

0 Members and 1 Guest are viewing this topic.

Offline izzyrocket

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Help on problems dealing with Percent Yields
« on: October 23, 2008, 09:40:04 AM »
these are problems for my intro to chem for engineers homework that i am having trouble with:

the first one i got the percent yield right, but i dont know how to do the other part. the two answers at the bottom are wrong.
http://i95.photobucket.com/albums/l129/wutaboutfitch/Picture2.png

for the second one i have no idea.
http://i95.photobucket.com/albums/l129/wutaboutfitch/Picture3.png


thank you!

Offline Astrokel

  • Full Member
  • ****
  • Posts: 989
  • Mole Snacks: +65/-10
  • Gender: Male
Re: Help on problems dealing with Percent Yields
« Reply #1 on: October 23, 2008, 09:53:40 AM »
1b - show your working?

2a - use percentage yield formula to get the answer as i think you can assume the 50.0g of ethanol did not produce any side reaction.

2b - how many moles of ethanol that did not converted to diethyl ether? And what is 45% of it?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline izzyrocket

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: Help on problems dealing with Percent Yields
« Reply #2 on: October 23, 2008, 12:07:48 PM »
1a: 83.2g Zn   1molZn     8molZnS   97.475 gZns
__________________________________________=123.9878<--- limiting reagent
                    65.409gZn 8molZn      1 mol ZnS                                                                                                   
104.9/123.9878*100=84.6%

but for part B the equation does the equation change? because it says oxygen is added to the reactants. i used the remaining amount of S (because all the Zn was used) to find how much ZnO and SO2 is made but i got it wrong

for number 2 i got that 50g of ethanol theoretically yields 40.223g so, 26.7/40.223 gives me 92% yield but i dont know how to do the second part

Offline Astrokel

  • Full Member
  • ****
  • Posts: 989
  • Mole Snacks: +65/-10
  • Gender: Male
Re: Help on problems dealing with Percent Yields
« Reply #3 on: October 23, 2008, 12:32:01 PM »
Quote
but for part B the equation does the equation change? because it says oxygen is added to the reactants. i used the remaining amount of S (because all the Zn was used) to find how much ZnO and SO2 is made but i got it wrong

yes the equation will change(zinc reacts with oxygen and sulphur reacts with oxygen). The reason why your percentage yield isn't 100% is because some amount of Zn has reacted with oxygen to formed ZnO. On top of the excess sulphur, also consider your amount of S as not all of the S supposed to react with Zn has reacted.

2b - 92% is correct, 2b is the same as 1b
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Sponsored Links