Equilibria, Energetics and Elements (F325) - June 2011 Exam.

Do you want the calculation specifically or?

Just did the January 2011 Paper. It wasn't too bad except the odd question about planning an experiment for working the enthalpy change of neutralisation and I didn't get the Q7 calculation AT ALL, well the first part was ok by working out the moles but thats about where I got to

Can anyone help me with that, at the same time explaining whats going on? Thanks

just everything lol !

Okay I'll run through the calculation:

As standard with these calculations, get anything you can into moles that's your first port of call:

So we know from the question we can calculate moles of Na2S2O3:
0.0246 X 0.001 = 0.0000246 moles

Now the difficult part:

You have to look carefully at the ratios, we have the moles of 2S2O3 so we know that this reacts with one mole of I2

Then look at the equation above, you see that One mole of I2 is equal to 2 moles of Mn(OH)3 Therefore the moles we have calculated is equal to the moles of Mn(OH)3

Naturally from this point you would like to X your moles by 2 and then divide by 4 DON'T DO THIS
What you have to remember is that your reaction mixture is the same regardless, an analogy is:
If you have a cup of 10ml of squash (undiluted) in one cup and you want to put it in a different cup it doesn't matter how much water you add, your initial quantity of squash is going to be the same regardless.

Same applies here, essentially what you're doing is taking you moles of Mn(OH)3 is taking them to a different reaction mixture where it reacts on a different ratio to the previous mixture your moles of Mn(OH)3 is going to be the same.

So looking at the top equation you know that you moles of Mn(OH)3 is 4 times the moles of O2 so divide the moles you have by 4. = 0.00000615 moles
Conc = 0.00000615/0.025= 0.000246 dm^-3
x 32 for g dm-3 = 0.007872 gdm-3
x1000 for mg dm-3 = 7.87mg -3

Good news everyone, 7.87>5 the fish can survive

Anyway it's twenty to 12, i'm shattered, really hope this has helped!

Do we get these equations or are we supposed to learn them:

Mno4- + 5Fe2+ + 8H+ --> 5Fe3+ + 4H2O + Mn2+

2S2O3(2-) + I2 --> S4O62- + 2I-

Do we get these equations or are we supposed to learn them:

Mno4- + 5Fe2+ + 8H+ --> 5Fe3+ + 4H2O + Mn2+

2S2O3(2-) + I2 --> S4O62- + 2I-

Oh thats a quite clever calculation actually but you wont be able to see that straightaway. Thanks mate!

LEARN.

Are you learning any others? cos I dont know which else to learn

I thought it was pretty standard for those calculations, how do you do them?

Yeah it is standard after you get the mid part, thats the most difficult part I reckon. The bit before and the things you have to do after aren't that hard. Its just identifying that 4:1 ratio which I was referring to

Okay I'll run through the calculation:

As standard with these calculations, get anything you can into moles that's your first port of call:

So we know from the question we can calculate moles of Na2S2O3:
0.0246 X 0.001 = 0.0000246 moles

Now the difficult part:

You have to look carefully at the ratios, we have the moles of 2S2O3 so we know that this reacts with one mole of I2

Then look at the equation above, you see that One mole of I2 is equal to 2 moles of Mn(OH)3 Therefore the moles we have calculated is equal to the moles of Mn(OH)3

Naturally from this point you would like to X your moles by 2 and then divide by 4 DON'T DO THIS
What you have to remember is that your reaction mixture is the same regardless, an analogy is:
If you have a cup of 10ml of squash (undiluted) in one cup and you want to put it in a different cup it doesn't matter how much water you add, your initial quantity of squash is going to be the same regardless.

Same applies here, essentially what you're doing is taking you moles of Mn(OH)3 is taking them to a different reaction mixture where it reacts on a different ratio to the previous mixture your moles of Mn(OH)3 is going to be the same.

So looking at the top equation you know that you moles of Mn(OH)3 is 4 times the moles of O2 so divide the moles you have by 4. = 0.00000615 moles
Conc = 0.00000615/0.025= 0.000246 dm^-3
x 32 for g dm-3 = 0.007872 gdm-3
x1000 for mg dm-3 = 7.87mg -3

Good news everyone, 7.87>5 the fish can survive

Anyway it's twenty to 12, i'm shattered, really hope this has helped!

Oh yeah, it takes some practice the trouble is most of the past papers give you an equation where from one to the other the moles either half or double which isn't very difficult at all and you can get lucky, you just know with the new spec. they're not going to make it easy.

Hey, thanks for your help on this question, I was stuck on this as well the only thing is I still don't understand why you divide by 4? I'm not very good at the ratio part of titraton questions generally why don't you times by 2 then divide by 4?

Well personally I found it difficult because I didn't relate it to the previous equations at all cos I worked out the moles of S2O3 and i was like "what should i do now?". I finished like the old spec questions yesterday and none of them had that. So what would you recommend for me to revise? Not specifically calculations but like generally

Hey, thanks for your help on this question, I was stuck on this as well the only thing is I still don't understand why you divide by 4? I'm not very good at the ratio part of titraton questions generally why don't you times by 2 then divide by 4?

No worries at all

The thing that you have to remember is that is a titration, and that the Mn(OH)3 produced in the first equation is going to be used in equation 2,
so essentially you're taking the reaction mixture from the first reaction and taking it to the 2nd reaction so you're overall amount of moles is the same. However, it reacts in different ratios in the 2nd reaction, but this does not change the amount of moles you had originally,

do you sort of understand, if not i'll try and think of something else?

Could someone give me the answer for this question:

A piece of iron wire of mass 2.225 g was dissolved in dilute suphuric acid and the solution was made up to 250cm3. 25cm3 of this solution required 31cm3 of 0.0185moldm-3 potassium dichromate (VI) solution for oxidation. Calculate the percentage of iron in the iron wire?