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

1. (Original post by student777)
I'm sorry, I still don't get it! I understand what you're saying, but I can't see how it relates to the question...
the extent of my photoshopping ability
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2. (Original post by wilsea05)
0.031x0.0185=0.0005735mol of dichromate
the ratio of dichromate:Fe will be 1:3? not sure tbh haven't got the half equations
times that by 10=mol of Fe in 250cm3
mol x 55.85= mass of Fe in wire
divide that by 2.225 and x100 = percentage.
Thanks.
3. (Original post by haydyb123)
the extent of my photoshopping ability
Can I just say, I really appreciate the effort you make answering people's questions- thank you!

I think I get it now. It's sort of... The whole precipitate has reacted, which is 4 mols. But the second equation shows you the ratio for 2 mols, but really it's 4? It's the simplest ratio but not how it actually reacts.

What i've now done is multiply reactions 2 and 3 by 2 so that the numbers of Mn(OH)3 match up with the first equation. Now the equation is easier to follow. Is that right?
4. (Original post by student777)
Can I just say, I really appreciate the effort you make answering people's questions- thank you!

I think I get it now. It's sort of... The whole precipitate has reacted, which is 4 mols. But the second equation shows you the ratio for 2 mols, but really it's 4? It's the simplest ratio but not how it actually reacts.

What i've now done is multiply reactions 2 and 3 by 2 so that the numbers of Mn(OH)3 match up with the first equation. Now the equation is easier to follow. Is that right?
It's nothing seriously .

On your point of doubling up, I think you could get caught in a trap if equations don't go together, but if it works in this instance and it helps you see for this question only then great. But I really think you just have to remember that the moles you've calculated in one equation has to be the same in the other equation, as essentially, you're working backwards. Hence why these are sometimes called backwards titrations.
5. (Original post by haydyb123)
It's nothing seriously .

On your point of doubling up, I think you could get caught in a trap if equations don't go together, but if it works in this instance and it helps you see for this question only then great. But I really think you just have to remember that the moles you've calculated in one equation has to be the same in the other equation, as essentially, you're working backwards. Hence why these are sometimes called backwards titrations.
Yeah you explained that titration question really well, thanks a lot!
6. So guys,

The half equations show the point of view of the reactants yes? so when we 2 we get a full ionic equation, thus the cancelling of the electrons... so if they give us something like 2s203 (2-) + 2I- ------>I2+s406 (2-)
we have to split them up by looking at each reactant?

And one more thing, why does CaCl2 form, and not CaCl3? they have variable oxidation states so why cant we just get that?
7. Could someone explain to me how, on page 221 of the OCR Chemistry A2 textbook for the iron tablet example, how the molar mass of 55.8g mol^-1 was calculated? I'm beginning to think they've left out some information.
8. (Original post by Tphys)
Could someone explain to me how, on page 221 of the OCR Chemistry A2 textbook for the iron tablet example, how the molar mass of 55.8g mol^-1 was calculated? I'm beginning to think they've left out some information.
Periodic table
9. (Original post by INeedToRevise)
Periodic table
...I'm stupid. Thanks.
10. Okay guys, this is some seriously hard question *props to anyone who gets it*

A 6.6g of a fertiliser containing ammonium sulphate was boiled with Excess of NaOH. All of the Ammonia gas liberated was absorbed in 50.0cm^3 of 1moldm^3 HCl (IN EXCESS). The resulting solution was diluted to exactly 250cm using pure water. A 25.0cm^3 portion of this solution was titrated with 0.1moldm^3 NaOH solution. 20.0cm^3 of the NaOH was required for neutralisation. Calculate the percentage of Ammonium sulphate in the fertliser.

Equations: (NH4)2S04 +2NaOH-----> Na2S04 +2NH3+2H20
NH3 +HCl------------------> NH4CL
NaOH+HCl-----------------> NaCl +H20

(8marks)
11. (Original post by J DOT A)
Okay guys, this is some seriously hard question *props to anyone who gets it*

A 6.6g of a fertiliser containing ammonium sulphate was boiled with Excess of NaOH. All of the Ammonia gas liberated was absorbed in 50.0cm^3 of 1moldm^3 HCl (IN EXCESS). The resulting solution was diluted to exactly 250cm using pure water. A 25.0cm^3 portion of this solution was titrated with 0.1moldm^3 NaOH solution. 20.0cm^3 of the NaOH was required for neutralisation. Calculate the percentage of Ammonium sulphate in the fertliser.

Equations: (NH4)2S04 +2NaOH-----&gt; Na2S04 +2NH3+2H20
NH3 +HCl------------------&gt; NH4CL
NaOH+HCl-----------------&gt; NaCl +H20

(8marks)
I have no idea but I've got an answer Is it 40%??
12. (Original post by J DOT A)
Okay guys, this is some seriously hard question *props to anyone who gets it*

A 6.6g of a fertiliser containing ammonium sulphate was boiled with Excess of NaOH.
All of the Ammonia gas liberated was absorbed in 50.0cm^3 of 1moldm^3 HCl (IN EXCESS).
The resulting solution was diluted to exactly 250cm using pure water. A
25.0cm^3 portion of this solution was titrated with 0.1moldm^3 NaOH solution. 20.0cm^3 of the NaOH was required for neutralisation. Calculate the percentage of Ammonium sulphate in the fertliser.

Equations: (NH4)2S04 +2NaOH-----> Na2S04 +2NH3+2H20
NH3 +HCl------------------> NH4CL
NaOH+HCl-----------------> NaCl +H20

(8marks)
36+32+64=132

0.1x0.02=0.002moles of NaOH
0.002 moles of NH3 x10 = 0.02moles in the 250cm3
0.02/2=0.01moles
0.01x132=1.32g
1.32/6.6x100=20%

no idea if i'm right tbh
13. (Original post by wilsea05)
36+32+64=132

0.1x0.02=0.002moles of NaOH
0.002 moles of NH3 x10 = 0.02moles in the 250cm3
0.02/2=0.01moles
0.01x132=1.32g
1.32/6.6x100=20%

no idea if i'm right tbh
*you missed one important step*
It was also in the practical assesments in the OCR evaluative task I think...

THE HCl is IN EXCESS.
So the titration you get is not the actual amount as it is in excess. As you can see we HAVE 2 VALUES of HCl, and HCl appears twice in the equations. It is in excess....
So what would you subtract?
14. (Original post by sportycricketer)
I have no idea but I've got an answer Is it 40%??
HCl is in EXCESS.
We have 2 calculations for HCl, one in the beggining and one at the end....
15. we did F325 in january before F324 O_o apparently coz we could re-sit it in june
16. (Original post by J DOT A)
HCl is in EXCESS.
We have 2 calculations for HCl, one in the beggining and one at the end....
60%
17. (Original post by J DOT A)
HCl is in EXCESS.
We have 2 calculations for HCl, one in the beggining and one at the end....
80%?
18. (Original post by wilsea05)
36+32+64=132

0.1x0.02=0.002moles of NaOH
0.002 moles of NH3 x10 = 0.02moles in the 250cm3
0.02/2=0.01moles
0.01x132=1.32g
1.32/6.6x100=20%

no idea if i'm right tbh
dont u need to account for the HCl that neutralises the NaOH im confused coz i got
0.002mol NaOH
and 0.005mol of HCl in 25cm3 of solution ? this means all the NaOH would be neutralised by the HCl can someone show me what im doin wrong?
19. (Original post by J DOT A)
HCl is in EXCESS.
We have 2 calculations for HCl, one in the beggining and one at the end....
what is the answerrrrrrrrrrrrrrrrrrrrrrr?!
20. (Original post by wilsea05)
what is the answerrrrrrrrrrrrrrrrrrrrrrr?!
The answer is 30% guys.

We have HCl twice.... so we need to take away one from the other.

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