Chemistry stiochiometry

SuperRobotWars

Hey,
I've been having trouble with stiochiometry for too long and have decided to remedy that. Below there should a pic of question on a mock paper and I was wondering if someone could check if its right or if its not give me some help, it would be greatly appreciated.
The link to the work is here: https://www.flickr.com/photos/123626386@N07/with/13980731003/

Mr Pseudonym

Haven't done Chem in a year or two, so I'm incredibly rusty. But, I'll give a few pointers to get the thread started! First, when you divided both masses by the molar masses, the answer is in moles, not grammes.

You found that ethanol is in excess, so it can effectively be ignore henceforth. One mole of sodium dichromate reacts (with the ethanol) to produce three moles of ethanal. Therefore, multiply number of moles of sodium dichromate by three. That gives you the number of moles of ethanal; multiply that by the molar mass of ethanal to establish the mass in grammes; and then you're home and dry.

As I said, I'm verrry rusty!

little sis...

Mr Pseudonym pretty much covered it all despite claiming to be rusty :P

Yeh the answer to the first part is right but the second is wrong.
You need to use the limiting reactant to find the amount of all other products, so in this case it's the sodium dichromate. You used the ethanol to calculate the yield of ethanal (with an Mr of 44 )but you can't because it is in excess so it doesn't give you the correct yield of ethanal.
Try it again and let me know your answer I'll see if it matches mine:)

Ompala

Mr Pseudonym wrote: »

Haven't done Chem in a year or two, so I'm incredibly rusty. But, I'll give a few pointers to get the thread started! First, when you divided both masses by the molar masses, the answer is in moles, not grammes.

You found that ethanol is in excess, so it can effectively be ignore henceforth. One mole of sodium dichromate reacts to produce three moles of ethanol. Therefore, multiply number of moles of sodium dichromate by three. That gives you the number of moles of ethanol; multiply that by the molar mass of ethanol; and then you're home and dry.

As I said, I'm verrry rusty!

You mean multiply by the number of moles of ethanal (which in this case they are both 3 anyway), not ethanol. Similarly for finding yield multiply by the molar mass of ethanal not ethanol, because ethanal is what you are producing.

Mr Pseudonym

Ompala wrote: »

You mean multiply by the number of moles of ethanal (which in this case they are both 3 anyway), not ethanol. Similarly for finding yield multiply by the molar mass of ethanal not ethanol, because ethanal is what you are producing.

My bad! I meant ethanal! I promise that was a typo!!

little sis... wrote: »

Mr Pseudonym pretty much covered it all despite claiming to be rusty :P

Told you I was rusty!

Ompala

Mr Pseudonym wrote: »

My bad! I meant ethanal! I promise that was a typo!!

You are grand, they are easy to mix up
Also OP you should know from the experiment section that dichromate will be limiting reagent for producing ethanal.
Likewise ethanol will be the limiting reagent when you are producing ethanoic acid.

SuperRobotWars

Sorry for the delay. Did the second part again hopefully its right this time round
https://www.flickr.com/photos/123626386@N07/13988433324/

little sis...

SuperRobotWars wrote: »

Sorry for the delay. Did the second part again hopefully its right this time round
https://www.flickr.com/photos/123626386@N07/13988433324/

perfect!