For complete combustion of methanol it is:
3 CH3OH + 2 H20 -> 2 CO2 + 4 H2O(apologies for the lack of subscript)
For complete combustion of methanol it is:
3 CH3OH + 2 H20 -> 2 CO2 + 4 H2O(apologies for the lack of subscript)
You don't need free oxygen?
And why are there:
3 carbons on the left and 2 on the right? 5 oxygen on the left and 8 on the right? 14 hydrogen on the left and 8 on the right?Summat not quite right there some where...
Anything that combusts without free oxygen, is an explosive - it will go bang all by itself. I'll remember that as I drink my next G&T (a mixture of ethanol & water rather than methanol & water, but what the hell. Cheers!).
I think Our Fred needs to revisit his O-level Chemistry.
I gave up chemistry after O level but even I can see you have a lot more to apologise for.
MBQ
er, no. It's
CH3OH + O2 -> CO2 + 2H2O
Still not right.
Each methanol needs two oxygen atoms for the carbon atom and two for the four hydrogen atoms. so that's 4 atoms in all, of which it can supply one. So it needs 3 extras. Well they come in pairs so its
2 x CH3OH + 3 x O2 -> 2 x CO2 + 4 x H2OGood thing none of you have a vote on energy policy or something..
Still doesn't balance. Try:
2 CH3OH + 3 02 -> 2 CO2 + 4 H2O
How embarrassing! Sorry, for typing nonsense. The above is what I meant to type. How I managed to put water on both sides of the equation I do not know! I wasn't supposed to be on Usenet when I posted it so perhaps it was because I was rushing before anyone spotted me! Sorry.
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