August 15, 2022, 03:05:41 AM
Forum Rules: Read This Before Posting

### Topic: Problem of the week - 09/07/2012  (Read 28239 times)

0 Members and 1 Guest are viewing this topic.

#### Rutherford

• Sr. Member
• Posts: 1868
• Mole Snacks: +60/-29
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #15 on: July 25, 2012, 03:24:57 PM »
Is it potassium ferrocyanide?

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #16 on: July 25, 2012, 06:33:00 PM »
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
If you're not part of the solution, then you're part of the precipitate

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #17 on: July 25, 2012, 06:36:39 PM »
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.
If you're not part of the solution, then you're part of the precipitate

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #18 on: July 25, 2012, 08:19:59 PM »
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?  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  2FeCl3 + 3H2O

FeCl3 + Cu0  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..
If you're not part of the solution, then you're part of the precipitate

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #19 on: July 26, 2012, 01:12:07 AM »
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?
If you're not part of the solution, then you're part of the precipitate

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #20 on: July 26, 2012, 09:44:58 AM »
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.
If you're not part of the solution, then you're part of the precipitate

#### Ann1234

• Regular Member
• Posts: 63
• Mole Snacks: +4/-0
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #21 on: July 26, 2012, 02:56:03 PM »
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..
« Last Edit: July 26, 2012, 03:07:39 PM by Ann1234 »

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #22 on: July 26, 2012, 04:10:01 PM »
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.

If you're not part of the solution, then you're part of the precipitate

#### Ann1234

• Regular Member
• Posts: 63
• Mole Snacks: +4/-0
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #23 on: July 26, 2012, 04:50:28 PM »
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...

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #24 on: July 26, 2012, 04:57:44 PM »
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 ..
If you're not part of the solution, then you're part of the precipitate

#### Ann1234

• Regular Member
• Posts: 63
• Mole Snacks: +4/-0
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #25 on: July 26, 2012, 06:20:45 PM »
um I think now I am on the right track...I'll post what I found a little later...working on it

#### Ann1234

• Regular Member
• Posts: 63
• Mole Snacks: +4/-0
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #26 on: July 26, 2012, 07:01:26 PM »
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

Finally...I think I got it

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)

« Last Edit: July 26, 2012, 07:52:02 PM by Ann1234 »

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #27 on: July 26, 2012, 08:02:22 PM »
Ann,
Great job!!!!!!!!!!!!!!!!!!!!!!!!

That is so simple in hindsight, not even thinking of Iron(IV).

Awesome job!!

High Five
If you're not part of the solution, then you're part of the precipitate

#### Ann1234

• Regular Member
• Posts: 63
• Mole Snacks: +4/-0
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #28 on: July 26, 2012, 08:09:42 PM »
Ann,
Great job!!!!!!!!!!!!!!!!!!!!!!!!

That is so simple in hindsight, not even thinking of Iron(IV).

Awesome job!!

High Five

thank you   our brainstorming helped a lot

#### AlphaScent

• Full Member
• Posts: 638
• Mole Snacks: +24/-7
• Gender:
##### Re: Problem of the week - 09/07/2012
« Reply #29 on: July 26, 2012, 08:12:45 PM »
2 Brains are always better than 1

If you're not part of the solution, then you're part of the precipitate