U5-June-2009-old-S
Posted by: intechemistry on: September 16, 2010
Unit 5, June 2009, “OLD syllabus” to be discussed here.
3 Responses to "U5-June-2009-old-S"
February 16, 2012 at 6:10 pm
thank u so much sir..you’ve put so much effort to answer my question..really appreciate it.
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February 16, 2012 at 1:00 am
To Baihaqi 11M1
Re U5, Q1 d (i) and Iii)
Moles of R = (mass of R / molar mass of R) = 1.98/198 = 0.01 moles of R
Moles of Fe(II)SO4 oxidised = (mass of Fe(II)SO4 / molar mass of Fe(II)SO4) = 4.56/152 = 0.03 moles of Fe(II)SO4
0.03 moles of Fe(II)SO4 contains 0,03 moles of Fe(II)
For Fe in Fe(II)SO4 to change into Fe3+, therfore 0.03 moles of electrons were lost.
These e- must have gone to R.
ASSUMING only 1 atom of Fe per molecule of R:
0.01 moles of Fe in R must have gained th 0.03 moles of electrons from the Fe(II)SO4, i.e. 3e- gained for each Fe. As the FINAL oxidation state was Fe3+ and the Fe in R gainsed three e- to get to Fe3+, the original oxidation state of Fr in R must have been +6
Comment: We could have assumed R contained three Fe’s If we did The OS of Fe in R would work out to +(!V). We could have assumed two Fe’s in R, or four or five. This question is a bit lousy in this respect. You need to make the assumption or spend a lot of time trying to figure out how to finish the question even though there no information – so you will get stuck. I think the info in part (ii) should have been given in this part, but it wasn’t
There is a LESSON here: Read the WHOLE question first before answering. But it is unusual for this to happen. Usually you are given enough info (up to and including) in the question you are doing, to be able to give an answer.
This assumption is born out in part (ii).
Mass of 1 Fe = 56
56 + 38x + 16y = 198
38x + 16y = 198-56
38x + 16y = 142
Other info: K will give +1 O.S.
and oxygen -2 Need to balance up a +6 charge on Fe
therefore must have more than three oxygens’s (or when Fe is added to K, we will end up with a +’ve charge remaining)
Lets choose 4oxygens’s just to get a handle. we only have 142 mass units to play with also.
16×4 = 64. So 142-64 = 78 that fits nicely with two K+ ions. So proposed formula = K2FeO4
Quite a lot of thought/work needed for just 1 mark and a bit too ‘trial and error’-ish for my liking.
Anyone got a better way to do it?