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AQA 2010 ISA Paper Q10

Hey, I am struggling with a question I encountered in a 2010 ISA paper. Here is a link to the paper:

http://filestore.aqa.org.uk/subjects/AQA-CHM6T-Q10-TEST.PDF

Here is a link to the mark scheme:

http://a-levelchemistry.co.uk/aqa%20chemistry/aqa%20a2%20chemistry/unit%206/isa%20(empa)/chm6t-q10-mg.pdf

The question is 12e(ii): Deduce the number of moles of iron (ii) ethanedioate that would react with one mole of potassium manganate (VII) in acidic solution.

The answer in the mark scheme is 1.67. I have no idea how they arrived at this answer.

Any help would be appreciated.
Original post by tbh12
Hey, I am struggling with a question I encountered in a 2010 ISA paper. Here is a link to the paper:

http://filestore.aqa.org.uk/subjects/AQA-CHM6T-Q10-TEST.PDF

Here is a link to the mark scheme:

http://a-levelchemistry.co.uk/aqa%20chemistry/aqa%20a2%20chemistry/unit%206/isa%20(empa)/chm6t-q10-mg.pdf

The question is 12e(ii): Deduce the number of moles of iron (ii) ethanedioate that would react with one mole of potassium manganate (VII) in acidic solution.

The answer in the mark scheme is 1.67. I have no idea how they arrived at this answer.

Any help would be appreciated.


Question 2 gives you the ratio of manganate(VII) to ethandioate ions and question 12b gives you the reaction ratio between ethandioate and iron(III).

The 12e(ii) asks for the reaction ratio between iron(II) ethandioate and manganate(VII)
Reply 2
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tbh12
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Original post by charco
Question 2 gives you the ratio of manganate(VII) to ethandioate ions and question 12b gives you the reaction ratio between ethandioate and iron(III).

The 12e(ii) asks for the reaction ratio between iron(II) ethandioate and manganate(VII)


Ok, So the ratios I have calculated are as follows:

Manganate (VII) 2: 5 Ethanedioate
Iron (II) 2: 1 Ethanedioate

..I have no idea how to use this information to calculate the ratio between Iron (II) ethandioate and manganate (VII) :-(
Original post by tbh12
Ok, So the ratios I have calculated are as follows:

Manganate (VII) 2: 5 Ethanedioate
Iron (II) 2: 1 Ethanedioate

..I have no idea how to use this information to calculate the ratio between Iron (II) ethandioate and manganate (VII) :-(


manganate(VII) oxidises ethandioate in a 5:2 ratio
iron(III) oxidises ethandioate in a 1:2 ratio

Therefore iron(II) to manganate(VII) is a 1:5 ratio
Reply 4
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tbh12
OP
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Original post by charco
manganate(VII) oxidises ethandioate in a 5:2 ratio
iron(III) oxidises ethandioate in a 1:2 ratio

Therefore iron(II) to manganate(VII) is a 1:5 ratio


Ah ok, yes I did get this far but dismissed it because I thought it was wrong. From this I calculate that if we have 1 mole of manganate(VII) this would react with 1/5 moles of iron(II) which is 0.2 However, the answer is 1.67. Sorry, I am probably being very stupid but I can't see where I am going wrong.
Original post by tbh12
Ah ok, yes I did get this far but dismissed it because I thought it was wrong. From this I calculate that if we have 1 mole of manganate(VII) this would react with 1/5 moles of iron(II) which is 0.2 However, the answer is 1.67. Sorry, I am probably being very stupid but I can't see where I am going wrong.


This is not what the question actually asks ...
Reply 6
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tbh12
OP
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Original post by charco
This is not what the question actually asks ...


hmm I really don't understand what is going on so you might have to spell it out to me. The question asks for the reaction ratio between iron(ii) ethanedioate and potassium manganate(VII) in acidic solution. The answer given is 1.67, or a ratio of Iron (II) ethandioate 5: 3 manganate (VII). I don't know how to get to this answer. Also, I don't know how the previous ratio you calculated of iron(II) 1: 5 manganate(VII) helps.
Reply 7
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tbh12
OP
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Anyone?
Original post by tbh12
hmm I really don't understand what is going on so you might have to spell it out to me. The question asks for the reaction ratio between iron(ii) ethanedioate and potassium manganate(VII) in acidic solution. The answer given is 1.67, or a ratio of Iron (II) ethandioate 5: 3 manganate (VII). I don't know how to get to this answer. Also, I don't know how the previous ratio you calculated of iron(II) 1: 5 manganate(VII) helps.


OK 12e(ii) Deduce the number of moles of iron(II) ethanedioate that would react with one mole of potassium manganate(VII) in acidic solution.

The equations:

Fe2+ --> Fe3+ + 1e
MnO4- + 8H+ + 5e --> Mn2+ + 4H2O
(C2O4)2- --> 2CO2 + 2e

Manganate(VII) absorbs 5 electrons
Iron ethandioate releases 3 electrons

Hence 1 mol manganate(VII) (5 electrons) will react with 1.67 mol iron(II) ethandioate (1.67 x 3 = 5 electrons)
Reply 9
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tbh12
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Ah I see, thanks a lot.

That seems a lot of work for just 1 mark!

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