Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: xstrae on September 25, 2006, 01:24:43 PM
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Maybe I am just losing it, but I am not able to get this one.
C2H5OH + I2 + OH- --> CHI3 + HCOO- + H2O
oxid states of C in C2H5OH : -3,-1
which of these do I consider or should I just take the average?
oxidation state of I in CHI3 : -1
and the oxidation states of both the C atoms in the products are +2
so should I take 4 oxidation reactions? I am confused.
Please help.
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You are being tricked, it can't be balanced without using negative coefficients.
Exchange OH- and HCOO- and try then. Forget about redox - just balance by inspection.
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Use a mean oxidation state for C in C2H5OH (-2)
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1) A 6.90 M solution of aqueous KOH has 30% by weight of KOH. Calculate the density of the solution.
Here is my work:
1 litre of the solution must contain 300ml of KOH
6.90 = n.of moles/.3
no. of moles of KOH = 2.07
wt of KOH = 117.99
density of solution = 117.99g/l
Where am I going wrong? My book says the answer is 1.288g/ml
2) What is the equivalent weight of FeC2O4 in the following reaction?
FeC2O4 ---> Fe+3 + CO2
MW = 144
oxid state of Fe in FeC2O4 is +2
oxid state of Fe+3 ion is +3.
eq. wt = MW/change in oxidation state = 144g
Book says the answer is 48g.
thanks for your time.
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1 litre of the solution must contain 300ml of KOH
No. 1 kg of solution must contain 300g of KOH.
http://www.chembuddy.com/?left=concentration&right=percentage-to-molarity
FeC2O4 ---> Fe+3 + CO2
Is this reaction balanced?
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ok I understood my mistake in the first problem.
FeC2O4 ---> Fe+3 + 2CO2 + 3e-
Fe loses one moles of electrons. C loses 2 moles of electrons.
so total change in oxidation state is 3.
eq wt = 144/3 = 48g.
thanks for your help.
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ok I understood my mistake in the first problem.
FeC2O4 ---> Fe+3 + 2CO2 + 3e-
Fe loses one moles of electrons. C loses 2 moles of electrons.
so total change in oxidation state is 3.
eq wt = 144/3 = 48g.
thanks for your help.
You are still wrong (iron wont get oxidized here, it gets reduced to Fe). But to be honest I have no idea what the correct answer is. Either brain fart or the question doesn't make sense.
Anybody?
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With normal methods I got,
5C2H5OH + 12I2 + 35OH- --> 8CHI3 + 2HCOO- + 17H2O i cant get the charges right though.
Exchange OH- and HCOO- and try then. Forget about redox - just balance by inspection.
how did you know this?
after trying to balance it after exchanging I got,
3C2H5OH + 12I2 + 2HCOO---> 8CHI3 + 2HCOO- + 5H2O + 2OH-
my book says the correct answer is
C2H5OH + 4I2 + 6OH- --> CHI3 + HCOO- + 5I- + 5H2O
how did you know that OH- and HCOO- had to be exchanged? And why arent redox rules working for this one? I think they should because this question was in the "easy MCQs" section of my book. It usally doesnt contain any perplexing problems.
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You are still wrong (iron wont get oxidized here, it gets reduced to Fe).
Are you sure? If you are, then my entire understanding of this must be wrong.
This is what I thought,
Fe+2 + C2O42- --> FeC2O4
Fe+2 --> Fe+3 + e-
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With normal methods I got,
5C2H5OH + 12I2 + 35OH- --> 8CHI3 + 2HCOO- + 17H2O i cant get the charges right though.
If charges are not balanced equation is not balanced.
Exchange OH- and HCOO- and try then. Forget about redox - just balance by inspection.
how did you know this?
Balance using algebraic method (http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method) - some coefficients are negative. That's a clear sign that some reagents are on the wrong side.
after trying to balance it after exchanging I got,
3C2H5OH + 12I2 + 2HCOO---> 8CHI3 + 2HCOO- + 5H2O + 2OH-
HCOO- on both sides - so there is something wrong.
my book says the correct answer is
C2H5OH + 4I2 + 6OH- --> CHI3 + HCOO- + 5I- + 5H2O
There was no I- in your first post. You copied something wrong - either question, or answer.
how did you know that OH- and HCOO- had to be exchanged? And why arent redox rules working for this one? I think they should because this question was in the "easy MCQs" section of my book. It usally doesnt contain any perplexing problems.
There are two independent reactions here:
C2H5OH + 4I2 + 10OH- -> 2HCOO- + 8I- + 7H2O
C2H5OH + 4I2 + 2OH- -> 2CHI3 + 2I- + 3H2O
Check out this equation balancing failures (http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure) lecture.
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Quote
after trying to balance it after exchanging I got,
3C2H5OH + 12I2 + 2HCOO---> 8CHI3 + 2HCOO- + 5H2O + 2OH-
HCOO- on both sides - so there is something wrong.
hey actually, I typed it wrong. there was no HCOO- on the products side. After exchanging and balancing I get, 3C2H5OH + 12I2 + 2HCOO- --> 8CHI3 + 5H2O + 2OH-
which is balanced. You were right, thanks.
my book says the correct answer is
C2H5OH + 4I2 + 6OH- --> CHI3 + HCOO- + 5I- + 5H2O
There was no I- in your first post. You copied something wrong - either question, or answer.
I posted exactly what was given. So I suppose the question is wrong then.
By the way, I have a doubt. It might be really silly, but anyway..
Say you are given an equation in which either H2O or OH- is present in the products or reactants. While balancing, is it important to take these into consideration while writing the individual oxidation/reduction reactions?
thanks for your time.