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

Reply 360

haydyb123

9

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 :wink:

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.

Reply 362

student777

8

Original post by haydyb123

the extent of my photoshopping ability :wink:

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?

Reply 363

haydyb123

9

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 :smile:

.

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. :smile:

Reply 364

k256_93

Original post by haydyb123

It's nothing seriously :smile:

.

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. :smile:

Yeah you explained that titration question really well, thanks a lot! :smile:

Reply 365

J DOT A

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?

Reply 366

Tphys

1

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.

Reply 367

INeedToRevise

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 :confused:

Reply 368

Tphys

1

Original post by INeedToRevise

Periodic table :confused:

...I'm stupid. Thanks.

Reply 369

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)

(edited 12 years ago)

Reply 370

sportycricketer

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)

I have no idea but I've got an answer :s-smilie:

Is it 40%??

Reply 371

wilsea05

1

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

Reply 372

J DOT A

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?

Reply 373

J DOT A

Original post by sportycricketer

I have no idea but I've got an answer :s-smilie:

Is it 40%??

HCl is in EXCESS.
We have 2 calculations for HCl, one in the beggining and one at the end....

Reply 374

wiggle-bacon

we did F325 in january before F324 O_o apparently coz we could re-sit it in june

Reply 375

sportycricketer

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%

(edited 12 years ago)

Reply 376

wilsea05

1

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%?

Reply 377

sohinder

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?