November 29, 2021, 05:03:09 AM
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Topic: Please calculate the number of mols of oxygen gas produced per second  (Read 58958 times)

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Offline Frederick95

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Please calculate the number of mols of oxygen gas produced per second
Remember PV=nRT and that you must account for the partial pressure of water vapour.

The reactions conditions are:
Atmospheric pressue:102.6 kPa;
Temperature:290 K
Partial pressure of water vapour:1.94 kPa.

The reaction is:
Volume of 6% H2O2 (mL):10.0
Volume of distilled water (mL):0.0
Volume of 1.0 M NaI (mL):10.0
Volume of distilled water (mL):0.0
O2 production in 60 s (mL):151.0
 :)

What I have done thus far is rearranged the Ideal Gas Law in order to get the number of mols

n = PV/RT

Is this the correct first step ???

Also, I have substituted the following values into the equation:
P = 102.6 kPa
V = ?
R = 8.3145 J/mol K
T = 290 K

 ???

Thank you for any help it is greatly appreciated.

Offline Arkcon

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O2 production in 60 s (mL):151.0

Did you misunderstand this parameter, for a start?


P = 102.6 kPa


Yeah that's what they give you, but maybe they've modified it in some way?

Let's see a balanced equation for the reaction first, the answer to a balanced equation is the moles, after all.
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Offline Frederick95

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The balanced equation would be

2H2O2 + NaI = 2H2O + O2 + Na + I

Am I correct, Arkcon?

Offline sjb

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The balanced equation would be

2H2O2 + NaI = 2H2O + O2 + Na + I

Does that make chemical sense? Sodium metal in the vicinity of water? Monoatomic iodine?

S

Offline Frederick95

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It make sense to me since both sides have equal quantities of elements/compounds.

Offline Borek

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It make sense to me since both sides have equal quantities of elements/compounds.

Just because it seems to be balanced doesn't mean it makes sense. Reread sjb post. What happens when you throw piece of sodium into water? You have sodium and water on the RHS of the equation - is it possible?
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Offline Frederick95

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Hmmmm....Sodium is a highly reactive metal, which breaks down water into hydrogen and oxygen when it comes in contact with it. Hence, the balanced equation should be:

2H202 + NaI = NaOH + I +H20?

I am not sure...

Offline Arkcon

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Elemental iodine exists as a diatomic molecule I2.  Also, isnt this reaction, as per the original question, supposed to form oxygen gas?
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Offline Frederick95

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I beleive I have it figured out:


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

Offline Arkcon

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I beleive I have it figured out:


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

Great.  A little Googling on my part confirms that NaI is just a catalyst.  (i hope that's true, Google could be wrong you know)

Now, how much O2 do you get?
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Offline Frederick95

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It appears we get 1 molecule of O2

Offline Frederick95

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What do I need to know the number of O2 molecules for? :-\

Offline Arkcon

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O2 production in 60 s (mL):151.0

Did you misunderstand this parameter, for a start?


Did you miss this, or did you not realize that mLs is a volume measurement, that you can plug into PV=nRT? 1ml=1cm3
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Offline Frederick95

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Sorry Arkcon, I forgot ::)
Since 1ml=1cm3

        151 mL = 151 cm^3

I beleive I now rearrange the Ideal Gas Law to find # of mols?

n = PV / RT

  =(102.6 kPa)(151 cm^3) / (8.3145 J/mol K)(290 K)

  = 15492.6 / 2411.205

  = 6.4 mols

 ???

Offline Arkcon

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Partial pressure of water vapour:1.94 kPa.


Now, what's this part for?
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