July 18, 2024, 06:51:30 AM
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Topic: Please calculate the number of mols of oxygen gas produced per second  (Read 65976 times)

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Offline Frederick95

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6.4 mols will tell us how many mols of O2 there are? Since there is one molecule of O2...

6.4 mols x 1 = 6.4 mols O2

Am I correct so far?

Now, since the oxygen production takes place in a time period of 60 seconds, I divide 6.4 mols/ 60 s

which gives 0.11 mols O2/sec  :-\

Offline Frederick95

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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2)

Offline Arkcon

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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2)

Augh.  So close.  And yet so far.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Frederick95

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Since this is a pressure of the substance in the gas phase which is established at a particular temperature,

The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2), at 290 K?

Offline Arkcon

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Since this is a pressure of the substance in the gas phase which is established at a particular temperature,

OK

Quote
The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water ,

OK

Quote
which is oxygen (O2)

What, you say water is oxygen?  I mean, you've typed it, so in a sense, you've said it, but, would you ever say that?  To a large group of people?

Quote
at 290 K?

And this addition adds, what, to your previous statement?
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Offline Frederick95

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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is H20.

Sorry, oxygen cannot be water since it has no hydrogen to perform hydrogen bonding

Offline Frederick95

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The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is water(g)(H2)), at 290 K meaning that at 290 K, the state of water is a gas.

Offline Arkcon

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The partial pressure, of water vapor, is pertinent to the oxygen generated, exactly how.

The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is H20.

Sorry, oxygen cannot be water since it has no hydrogen to perform hydrogen bonding

The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is water(g)(H2)), at 290 K meaning that at 290 K, the state of water is a gas.

When you write semi-logical, technically incorrect, mish-mashes of facts like this, I begin to suspect I'm talking to a bot. :D
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Frederick95

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The partial pressure, of water vapor, is related to the oxygen generated because oxygen in water obeys Henry's law rather well; the solubility is roughly proportional to the partial pressure of oxygen in the air:

pO2 = KO2 xO2

Offline Frederick95

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Was my statement correct? :-\

Offline Marcus Soutlo

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Arkcon, when you said

NaI + 2H2O2 --> NaI + 2H2O + O2

is correct, my teacher told me you were wrong.

Does anyone know what is the correct balanced equation?

Offline Arkcon

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Arkcon, when you said

NaI + 2H2O2 --> NaI + 2H2O + O2

is correct, my teacher told me you were wrong.

Does anyone know what is the correct balanced equation?


I think it's one correct equation, Googling leads me to believe that NaI is a catalyst for the decomposition of H2O2

Lookie here:

http://www.google.com/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=GFx&q=NaI+catalyst+hydrogen+peroxide&btnG=Search

Second one, I can't get the PDF to download, or I'd attach it for you.

Anyway, we did establish, I believe, that we won't be using the reaction, as volume of O2 was a given.
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Offline Borek

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I was taught to put catalyst over arrow, not on both reactants and products sides. I believe I have already wrote (in one of the numerous threads about this problem - you see now why posting the same zillion times doesn't make sense?) that NaI should be canceled.

Edit: even if - from the logical point of view - putting catalyst on both sides makes sense :)
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Offline Marcus Soutlo

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NaI + 2H2O2 --> NaI + 2H2O + O2

Cancelling NaI on both sides ;D

2H202 --> 2H20 + 02


But since the purpose of this experiment is to determine the order of the reaction with respect to hydrogen peroxide and iodide ion and to determine the rate law equation and the rate constant for the  decomposition of hydrogen peroxide catalyzed by iodide ion.

I am guessing that there would need to be an iodide ion present in the equation. So I came up with the following:

(1) H2O2 + 2NaI = 2Na+ + I2 + 2OH-
(2) H2O2 = H2O + 1/2 O2
Probably, (if pH is >7) there is also the formation of some IO3-:
3I2 + 6OH- = IO3- + 5I- + 3H2O

I got this information from http://www.thenakedscientists.com/forum/index.php?topic=2067.75;topicseen


Offline Borek

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(1) H2O2 + 2NaI = 2Na+ + I2 + 2OH-

Unless you will later reduce I2 to I- NaI is consumed, so it is hardly a catalyst.
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