RogueRose
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Calculating O2 produced from H2O2 (in variouos concentrations)
I thought it might be helpful for some people (and myself) if we could figure out how much O2 was produced from a specific concentration of H2O2. I
figured I'd set up an equation to calculate the amount of O2 with the given percentage/concentration of H2O2 in solution.
H2O2 = 34g/mole
O2 = 32g/mole
H2O = 18g/mole
2 mole = 1 mole + 2 mole
2H2O2 (68g)= O2 (32g) + 2H2O (36g)
O2 density (at STP) = 1.429g/L
O2
1ft^3 = 40.46g or 1.43 oz
1L = 1.43g or .05 oz
1lb = 12.076 standard cubic ft or .3174 Nm^3
1kg = 26.62 standard cubic ft or .6998 Nm^3
Here are a couple links to help calculate weights and volume of O2 for anyone who might need this:
Volume to weight conversion - shows weight of any given volume of pure O2 - calculator format
https://www.aqua-calc.com/calculate/volume-to-weight
Converts equal weights and volume equivalents of O2
http://www.airproducts.com/products/Gases/gas-facts/conversi...
1 standard ft^3 = 28.317 Liters
1 mole O2 = 32g : 1 L O2 = 1.429g
32g/1.429g= 22.39L O2 = 1 mole O2
Here is a page where the molarity of H2O2 is being calculated and I find it very difficult to follow how the % concentration translates to the numbers
that are used, for example the 30% listed H2O2 (weight/weight) he translates to 33.3% H2O2 solution (weight /volume) and comes up to 9.79 molarity
which I'm guessing is at 1L of 30% solution??.
https://www.scientistsolutions.com/forum/biochemistry-assay-...
If this is the correct molarity then a 35%w/w solution is 11.42 molarity and a 3% solution would be .979 molarity per liter.
35%w/w = 11.42 molarity per L
12%w/w = 3.916 molarity
3%w/w = .979 molarity
As the calculations above showed, 2mole H2O2 = 1 mole O2. So:
35% H2O2 = 11.42 moles = 5.71 moles O2
5.71 moles O2 (32g) = 182.72g O2
1L = 1.43g O2
182.72g O2 / 1.42g (per L) = 128.68 L or 4.544 ft^3
4.544 ft^3 O2 is equivalent to 21.64 ft^3 of air (at 21% O2 concentration)
1 gal of 35%w/w H2O2 = 484L O2 or 17.1 ft^3
1 Mole H2O2:
= .5 mole O2
1 mole O2 = 32g , .5mol = 16g
16g O2 / 1.43g (per liter) = 11.19L O2 per 1 mole H2O2
1 mole H2O2 yields 11.19L O2 gas at STP
Composition of Air:
Oxygen: 20.99%
Nitrogen: 78.03%
Carbon Dioxide: 0.03%
Hydrogen: 0.00005%
Argon: 0.93%
Neon: 0.0018%
Helium: 0.0005%
Krypton: 0.0001%
Xenon: 0.000009%
[Edited on 5-10-2018 by RogueRose]
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VSEPR_VOID
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Thanks. I have never need ti but that looks like a easy guild to follow. Have you considered writing a publication?
Within cells interlinked
Within cells interlinked
Within cells interlinked
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DraconicAcid
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Mood: Semi-victorious.
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3% H2O2 is known as "10 Volume" H2O2 because one L of it will decompose to give 10 L oxygen.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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AJKOER
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Calculated amount of O2 liberated from H2O2 and observed can vary more than one would expect. To estimate the latter, I would suggest reacting various
concentrations of H2O2 with excess chlorine bleach (all taken from the same bleach bottle in short order).
Compare the relative amount of oxygen generated per the equation:
NaOCl + H2O2 --> NaCl + H2O + O2 (g)
assuming for all concentrations of H2O2 tested, there was an excess of NaOCl employed.
You should repeat the experiment several times to address any measuring and gas collection issues.
I would also record the pH of the H2O2 (more alkaline solutions are more prone to decomposition and have a reduced shelf life). Note any listed
stabilizers also, all in an effort to explain any unexpected results.
[Edited on 10-5-2018 by AJKOER]
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RogueRose
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Registered: 16-6-2014
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Thanks for the compliment! I'm not really finished with the guide, I wrote it while juggling a few other things (copper experiment & family stuff)
and wanted to post what I had to get some feedback on it. I usually end up trying to consolidate some things, adding bold/underlines and making it
easier to follow & show some examples. I need to verify the molarity conversion which is the area right under the third link. I used data from
that site and the math looked correct but I need to work it though on paper to make sure.
I've thought about some publication stuff and have thought about doing a chemistry tutor site or a home school site because I have a lot to learn
though I have learned a lot on this site, it would be good to write things as I learn them because as the information is new, it is easier to
determine what topics need more focus, how to better explain them and how new terms fit together. There are some chemistry books that make it very
difficult to learn the subject because they just cover subjects by chapters instead of a nice flow of application and what each "piece" means, stands
for, or is used for.
*the word "math" came up as a mis-spelled word in my spell checker. It wasn't even in the dictionary, can you believe that!!!
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