[Frederick95]

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.

[Arkcon]

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.

[Frederick95]

The balanced equation would be

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

Am I correct, Arkcon?

[sjb]

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

[Frederick95]

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

[Borek]

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?

[Frederick95]

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...

[Arkcon]

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

[Frederick95]

I beleive I have it figured out:


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

[Arkcon]

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?

[Frederick95]

It appears we get 1 molecule of O2

[Frederick95]

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

[Arkcon]

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

[Frederick95]

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

 

[Arkcon]

Partial pressure of water vapour:1.94 kPa.

Now, what's this part for?