Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on July 09, 2012, 07:44:27 AM
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When 1.000 g of a strong oxidizer A was dissolved in water, solution initially became wine red, but soon the substance decomposed, producing 85 mL (STP) of a dry gas B and 0.4032 g of a red-brown solid C. Drop of a concentrated, oxidizing acid, added to the alkaline filtrate, produced white precipitate D. Red-brown solid C was dissolved in a hydrochloric acid, and 1 g copper wire was put into the solution. After reaction ended, mass of the wire left was 0.8396 g.
Name all substances, write all reaction equations.
This one is not very difficult, call it a side product of my current work on the stoichiometry calculator :P
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If I could think of a gass that has the molar mass 157.27g/mol.
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No idea where did you get 157 g/mol from?
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The number of moles of the gass B is 0.085/22.4=0.003795, and its mass is 1-0.4032=0.5968g. M=0.5968/0.003795=157g/mol.
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No, mass is off.
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I guess you mean that not the whole amount decomposed.
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Decomposition was stoichiometric and went to completion.
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Then why is the mass off?
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I see now that the wording is not perfect, but you ignored information about the filtrate.
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Oh yes, my mistake, sorry.
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Looks like week was not enough... Which most likely means nobody but Raderford even tried :-\
Deadline extended, I am going to wait for another week.
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Looks like week was not enough... Which most likely means nobody but Raderford even tried :-\
Deadline extended, I am going to wait for another week.
Borek, is not that we are not trying...I actually tried 2 times (last was yesterday) but I got stuck finding the strong oxidizer.... :(
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Borek, is not that we are not trying...I actually tried 2 times (last was yesterday) but I got stuck finding the strong oxidizer.... :(
Unfortunately I am away, so I can't answer immediately - but perhaps if you will post what you tried and what you found, it will either wait for a day or two till I will be able to comment, or it will trigger comments from others. I think there are things that should be obvious, like the identity of the red-brown solid.
OK, if someone is trying, I am extending it for another week.
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okay....I think the red brown solid may be Iron (III) Oxide, and I found that solutions of Fe(OH)3 are wine red...but this hydroxide in water seems to be stable...so I got stuck again :(
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Good idea about iron(III) - but it is not the original compound. Its a product of the original compound decomposition.
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Is it potassium ferrocyanide?
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I feel like saying oxidizing acid is too vague. Once you add said acid and substance D crashes out (precipitates), it is insolable in water. That has to be a major hint. But you don't specify how much crashes out? Does that mean we assume that it is 1 drop= 1mL of 37% HCL??
What we know:
Gas B = 3.79*10-3 moles
Assumptions:
Red-Brown Solid is Iron (III) Oxide, based upon rust's ability to dissolve in acid
Still working at it, but feel like more info needs to be given
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I apologize,
We also know that Red Brown solid C reacts to completion with copper, and 0.1604 g copper reacts. All reactions are stoichiometric; then there is 0.00252 moles of substance B after initial reaction.
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I do not know If any of this is right. When you look at a problem for long enough you start to second guess your self. I still need help in the beginning.
A? :rarrow: B? + C (Ferric Oxide) + alkaline filtrate
A= Ferric thiocyanate or ferric hydroxide
Alkaline solution + Oxidizing acid (HNO3, H2SO4, percholrate)?? :rarrow:D??
D= insoluble in water
C, Fe2O3 + 3 HCL :rarrow: 2FeCl3 + 3H2O
FeCl3 + Cu0 :rarrow: Fe2+ + Cu2+ + 3 Cl-
This may be wrong, but from Ferric Oxide to the Copper I know I am right. Its this beginning strong oxidizer to a gas and rust. Maybe iron is even involved???
Being at work late waiting on a reaction, its funny that in my down time trying to wrap my head around another reaction..
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Good idea about iron(III) - but it is not the original compound. Its a product of the original compound decomposition.
So from that we know it is a ferrous compound. It is then oxidized. So what does it reduce?
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When 1.000 g of a strong oxidizer A was dissolved in water
Good idea about iron(III) - but it is not the original compound. Its a product of the original compound decomposition.
These two statements contradict each other. Any ferrous compounds are reducing agents not, oxidizers. If Fe(III) is a product of the first decomposition then you have to start with Fe(III). It can be a strong oxidizer.
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I don't think A is a ferrous compound since - as far as I know -those tend to produce a green color, and this one produces a wine red color, characteristic of iron 3 solutions.
I am not sure but maybe Borek meant to say that Iron (III) Oxide was not the original compound but a product of the original compound decomposition (?), which I agree.
I think Iron (III) Oxide is C, the red-brown solid product of the decomposition of A. I also think A has to have Fe (III) to be a strong oxidizer agent.
Maybe Borek can clarify this quotation for us:
Good idea about iron(III) - but it is not the original compound. Its a product of the original compound decomposition.
I also thought on these two that you mentioned: A= Ferric thiocyanate or ferric hydroxide
But I am stuck since ferric hydroxide seems to be stable in solution, and with ferric thiocyanate I didn't find data about its stability.. :(
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Ann,
I agree with all of your points except Iron(III)hydroxide. That is basically hydrated Iron(III)oxide and is not very soluble in water. Ksp= 2.79*10^-39.
I honestly believe that there is something missing. It would work better if he gave us the amount of precipitate D.
???
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Ann,
I agree with all of your points except Iron(III)hydroxide. That is basically hydrated Iron(III)oxide and is not very soluble in water. Ksp= 2.79*10^-39.
I honestly believe that there is something missing. It would work better if he gave us the amount of precipitate D.
???
oh you're right ..and since that 1 gram was fully dissolved in water * I suppose at room temperature* it can't be iron (iii) oxide. I am thinking then on a salt, and it must be soluble...
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There really is not enough information to solvene the problem.
Ferrous Sulphate when heated releases SO2 and SO3 and leaves behind Iron(III)oxide. But it is not a strong oxidizing agent. Like you said, ferrous ions are blue to green. I am honestly going to say my answer is that the problem is unsolvable.
Ferric thiocyanate is perfectly stable material in water. It is used as fake blood in movies. For is to release hydrogen cyanide, it would take a strong acid and lots of heat. It is not Ferric thiocyanate.
I'm stumped .. :-X
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um I think now I am on the right track...I'll post what I found a little later...working on it :P
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When 1.000 g of a strong oxidizer A was dissolved in water, solution initially became wine red, but soon the substance decomposed, producing 85 mL (STP) of a dry gas B and 0.4032 g of a red-brown solid C. Drop of a concentrated, oxidizing acid, added to the alkaline filtrate, produced white precipitate D. Red-brown solid C was dissolved in a hydrochloric acid, and 1 g copper wire was put into the solution. After reaction ended, mass of the wire left was 0.8396 g.
Name all substances, write all reaction equations.
This one is not very difficult, call it a side product of my current work on the stoichiometry calculator :P
Finally...I think I got it ;D
When 1.000 g of a strong oxidizer A was dissolved in water, solution initially became wine red, but soon the substance decomposed, producing 85 mL (STP) of a dry gas B and 0.4032 g of a red-brown solid C.
4 K2FeO4(s) + 4 H2O(l) → 3 O2(g) + 2 Fe2O3(s) + 8 KOH(aq)
A = K2FeO4 (potassium ferrate, iron (VI) rare purple salt, very reactive with water)
B = O2
C = Fe2O3
And KOH gives the filtrate its alkalinity properties.
Drop of a concentrated, oxidizing acid, added to the alkaline filtrate, produced white precipitate D.
KOH(aq) + HClO4 (conc) = KCLO4(s) + H20(l)
D = KClO4 (white precipitate, lowest solubility of the alkali metal perchlorates)
Red-brown solid C was dissolved in a hydrochloric acid, and 1 g copper wire was put into the solution. After reaction ended, mass of the wire left was 0.8396 g.
Dissolution of C:
Fe2O3(s) + 6HCl(aq) = 2FeCl3(aq) + 3H20(l)
Redox reaction with copper wire:
2FeCl3(aq) + Cu^0(s) = CuCl2(aq) + 2FeCl2(aq)
:D
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Ann,
Great job!!!!!!!!!!!!!!!!!!!!!!!!
That is so simple in hindsight, not even thinking of Iron(IV).
Awesome job!!
High Five ;D
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Ann,
Great job!!!!!!!!!!!!!!!!!!!!!!!!
That is so simple in hindsight, not even thinking of Iron(IV).
Awesome job!!
High Five ;D
thank you :) our brainstorming helped a lot ;D
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2 Brains are always better than 1 ;)
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Ann,
Great job!!!!!!!!!!!!!!!!!!!!!!!!
That is so simple in hindsight, not even thinking of Iron(IV).
Awesome job!!
High Five ;D
To be correct - iron(VI) !
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Ha Ha, I meant that...my mistake 8)
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And that's the correct answer :D
Look at the attachment to see how all these numbers were obtained ;)