J
John Woodgate
Guest
I read in alt.binaries.schematics.electronic that Tim Williams
<tmoranwms@charter.net> wrote (in <3OhZd.54034$jl6.48778@fe06.lga>
about '~~Juice~~', on Mon, 14 Mar 2005:
The 'gas' in this case is H2O. K2O doesn't 'burn' in air but it *might*
attract more oxygen to form KO2.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
<tmoranwms@charter.net> wrote (in <3OhZd.54034$jl6.48778@fe06.lga>
about '~~Juice~~', on Mon, 14 Mar 2005:
Since my equation shows K2O being formed, I don't see what you doubt."John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote in message
news
It *might* happen to some extent because the hydrogen escapes and that
would tend to drive the reaction to the right. However, I suspect that
atmospheric oxygen (and perhaps even nitrogen) participates:
2KOH + Mg + O2 -> K2O + MgO + H2O
Doubtful to me, since gas is escaping all the while. I don't know if
K2O burns in air, but K metal (still hot) certainly does.
The 'gas' in this case is H2O. K2O doesn't 'burn' in air but it *might*
attract more oxygen to form KO2.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk