Question

Chemistry help on a Lab I did ?

Answer 1

it's likely that you wont get help if you don't post the actual question

Answer 2

Oh ops lol. 2. Below are the balanced equations for the reactions in Parts 1–3. Review the results of your experiments and look closely over the mole ratios of the chemical reactions. Do your experiments reflect the stoichiometric, molar amounts indicated by the reactions? Discuss this using the evidence from your experiments.
Cu(NO3)2 (aq) + 2NaOH (aq)  Cu(OH)2 (s) + 2NaNO3 (aq) Note: Cu(OH)2 is a blue precipitate.
FeSO4 (aq) + 2NaOH (aq)  Fe(OH)2 (s) + Na2SO4 (aq) Note: Fe(OH)2 is a dark green precipitate.
Fe(NO3)3 (aq) + 3NaOH (aq)  Fe(OH)3 (s) + 3NaNO3 (aq) Note: Fe(OH)3 is a red-orange precipitate



I took a screen shot of the Lab I worked on.

Answer 3

I dont understand how to see if my experiments reflect the stoichiometric, molar amounts indicated by the reactions?

Answer 4

This is my last question to do and I'm stuck on it :/ Can anyone explain to me or walk through with me on this ?

Answer 5

did they give you the concentrations they used for each well?

Answer 6

Oh yes!
Well # # Drops of Iron Sulfate # Drops of Sodium Hydroxide
A1 5 45
A2 10 40
A3 15 35
A4 20 30
A5 25 25
A6 30 20
B1 35 15
B2 40 10
B3 45 5

Answer 7

Thats for Microlpate A. I think Cu(NO3)2 (aq) + 2NaOH (aq)  Cu(OH)2 (s) + 2NaNO3 (aq) Note: Cu(OH)2 is a blue precipitate. Im not sure Im really confused.

Answer 8

no, that's for B, because it says "Iron Sulfate".

which is this equation:

FeSO4 (aq) + 2NaOH (aq)  Fe(OH)2 (s) + Na2SO4 (aq)

Answer 9

Omg yeah I just saw that.

Answer 10

I got them mixed up. Hold on let me get A and C.

Answer 11

wait let me just explain this one to you, you can work on the rest after

Answer 12

Yeah okay that would be great!

Answer 13

Im really confused.

Answer 14

FeSO4 (aq) + 2NaOH (aq)  Fe(OH)2 (s) + Na2SO4 (aq)

ratio of FeSO4:NaOH is 1:2, so you need to moles of NaOH for every mole of FeSO4.

Well # # Drops of FeSO4 # Drops of NaOH
A1 5 45 (ppt)
A2 10 40 (ppt)
A3 15 35 (ppt)
A4 20 30 (ppt)
A5 25 25 (ppt)
A6 30 20 (ppt)
B1 35 15 (almost no ppt)
B2 40 10 (no ppt)
B3 45 5 (no ppt)

A1-A6 had precipitate, though less ppt every time.
This corresponds to the decrease in the number of NaOH drops

At A2 10/1:40/2 the stoichiometric ratio is equal, so you should've seen the post ppt present. But looking at the picture it doesn't look right. I think you might've labelled them wrong, or there is an error somewhere. This fits microplate A better.

You should double check that.

Answer 15

Sorry i couldn't be of much help, the data and the results don't really make sense.

Answer 16

Oh gosh yeah im sorry I think I did do this wrong... hmm I better redo this Lab and Go over ore things. thank you so much though for trying to help. No wonder why I am confused I didn't do it right lol.