Capt Chaos
Harmless
Posts: 6
Registered: 2-1-2012
Member Is Offline
Mood: No Mood
|
|
H2O2 Decomp to O2 - Volume Calc?
Okay, I'm tired and need a peer review/conflict resolution.
What did I do wrong below?
40 volume peroxide is only supposed to produce 40 x 250ml = 10 L, not 21. I've even done the experiment and it's more like 10 L (maybe a little less).
Definitely not 21 L.
It's probably something obvious but I'm bleary. Have to iron this out tomorrow, so would love some input. I'll revisit tomorrow.
|
|
|
Poppy
Hazard to Others
Posts: 294
Registered: 3-11-2011
Member Is Offline
Mood: † chemical zombie
|
|
It's by definition, the volume of oxygen gas the peroxide would release if decomposed completly at room temperature. Thus 40 volumes would mean a
250mL bottle would release 40x250mL oxygen gas, thats 10L <_<
|
|
|
Neil
National Hazard
Posts: 556
Registered: 19-3-2008
Member Is Offline
Mood: No Mood
|
|
Looks like you mixed up O and O2 --> ~20L of O vs ~10L O2
|
|
|
Capt Chaos
Harmless
Posts: 6
Registered: 2-1-2012
Member Is Offline
Mood: No Mood
|
|
I am not sure I follow.
O2 is the correct in the formula and most of the references are fairly clear, stating that the volume rating is based on oxygen (O2).
It would seem unusual to assign a volume to just 'O'. Would O have a volume in this context and if it did, would it be likely to be half of O2?
|
|
|
unionised
International Hazard
Posts: 5126
Registered: 1-11-2003
Location: UK
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by Poppy  | | It's by definition, the volume of oxygen gas the peroxide would release if decomposed completly at room temperature. Thus 40 volumes would mean a
250mL bottle would release 40x250mL oxygen gas, thats 10L <_< |
No it isn't.
It's the volume you get if you add it to excess of an oxidant - typically KMnO4.
That way you get twice as much oxygen given off so your product looks twice as good.
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
