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Topic: Fe + O2  (Read 23457 times)

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Offline tortoise

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Fe + O2
« on: August 03, 2005, 09:41:51 AM »
what the product would be?

Offline sdekivit

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Re:Fe + O2
« Reply #1 on: August 03, 2005, 10:48:51 AM »
Fe2O3 i guess.

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Re:Fe + O2
« Reply #2 on: August 03, 2005, 06:31:34 PM »
Well, there are two oxides of iron: Fe2O3 and Fe3O4.  the first is just common rust, while the latter is called magnetite, a black, magnetic (duh!) substance.  While a rusting piece of iron tends to produce red rust; burning iron, in my experience, creates a product that is much darker; indicating that some magnetite might be formed as well.

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Re:Fe + O2
« Reply #3 on: August 03, 2005, 06:54:20 PM »
Well, there are two oxides of iron: Fe2O3 and Fe3O4.

Not exactly. The two oxides are Fe2O3 and FeO. Fe3O4 is a mixed oxide, you may treat it as Fe2O3.FeO.
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Offline tortoise

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Re:Fe + O2
« Reply #4 on: August 04, 2005, 09:21:04 AM »
The high school students who want to enter universities have to pass an important exam called entrance exam to universities in Vietnam. This year, this exam began on 3rd and finished on 10th of July.

In the exam this year, the chemistry exam (Grade A) had this problem:

III. 2. By burning comletely 33,4g the mixture B1 which consists of 3 metals Al, Fe & Cu in the atmosphere, we have 41,4g the mixture B2 which consists of 3 oxides. Then let B2 react comletely with the solution H2SO4 20%, d = 1.14 g/ml.
a) Write all the reaction equations.
b) Find the value of the minimum volume of H2SO4 20% which can mix the B2.

And the solution was given:
3Fe + 2O2 = Fe3O4.

But many students wrote Fe2O3. So they had no marks. This cause the protest from both the students and the teachers, but there was nothing new. The teachers of the Education and Training Department kept their solution.

After quoting from many believable books, a university teacher said that:
* 3Fe + 2O2 = Fe3O4 (at 150 - 6000C)
   The oxidization continues if O2 is odd:
   4Fe304 + O2 = 6Fe2O3 (at 450 - 6000C)
* That Fe reacts O2 produces the mixture: Fe2O3 + Fe304
* Therefore, either Fe2O3 or Fe3O4 had to be accepted.
« Last Edit: August 05, 2005, 08:30:03 AM by tortoise »

Offline tortoise

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Re:Fe + O2
« Reply #5 on: August 04, 2005, 09:35:51 AM »
That problem is a "funny" problem  ;D The data given is wrong ! Do you see that?

By culculating carefully, you will have nFe = 0.1875 (mol), nCu = 0.4 (mol) and nAl = -0.1 (mol) !
Actually, 27.(-0.1) + 56.(0.1875) + 64.(0.4) = 33.4g = mB1.

Hey, what do you think?
« Last Edit: August 05, 2005, 08:27:30 AM by tortoise »

Offline tortoise

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Re:Fe + O2
« Reply #6 on: August 09, 2005, 09:30:52 AM »
5 days has passed .

I am wondering if you don't concern to this matter...

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