# Complete Combustion of Butanol

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The millionth time youve helped me

thanx

cheers

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Anytime dude

and If Im not here, others will do same

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thats very ensuring.

As long as ur here, i dunno if a quiestion cannot be answered !!!!

Keep up the good work

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Ive never fathomed out how to open a tin can with a Banana yet, Im sure it has something to do with acidic properties though

lol

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First time to join the scienceforums, but first time to see such a rediculous debate. Why aommaster always counts the butanol as if there is 2 mole eq of butanol in the reactio? It just has 1 mole eq of butanol !!! well, of course you made mistake since your 1st post. (Bond E of butanol = 1 x 5580 !!!!)

C4H9OH + 6O2 ----------> 4CO2 + 5H2O

(tot E=8569.8) (tot E=11080)

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hey easy on the attacks, were all here to learn what we dont know........not everyone ca be as great as you!

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YT2095 said in post # :

theres 2 schools of thought about that in all honesty. for instance Phosphourous Pentoxide is often writen as P2O5 when in fact the molecule is P4O10.

That's different. ENTIRELY different.

If you've got a reaction, you should always cancel down numbers to their lowest possible integer states (occasionally, it's convention to express them as fractions as well, expressing the ratio to the combusting/reacting chemical (well, the one you're testing)), not doing it is wrong.

This is different from, say, cancelling H2O2 to HO. That's not a ratio of reactions.

If you had only one molecule of butanol, that ratio of oxygen atoms would still be sufficient to make it undergo complete combustion.

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yessipermana said in post # :

First time to join the scienceforums, but first time to see such a rediculous debate.

Be nice dude.

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MrL_JaKiri said in post # :

That's different. ENTIRELY different.

If you've got a reaction, you should always cancel down numbers to their lowest possible integer states (occasionally, it's convention to express them as fractions as well, expressing the ratio to the combusting/reacting chemical (well, the one you're testing)), not doing it is wrong.

This is different from, say, cancelling H2O2 to HO. That's not a ratio of reactions.

If you had only one molecule of butanol, that ratio of oxygen atoms would still be sufficient to make it undergo complete combustion.

LOL, that works probably fine for maths lessons or whatever, it is certainly NOT the way I was taught!

and youde certainly come unstuck in energetics if you tried that! heheheh.

Im sure there are SOME instances where you could apply that and get away with it, but to treat each molecule as its WHOLE is the way I was taught, same applies to dimetric or trimetric situations, you CANNOT afford to take to the lowest possible int, itll cock up all yer calcs else. its great if all elements(parts) have a single common denominator, and then factor up afterwards, but to avoid any mistakes you do it properly from the start.

btw, it took you a long time to suddenly come up with this out the blue? it has been a busy day for you

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YT2095 said in post # :

LOL, that works probably fine for maths lessons or whatever, it is certainly NOT the way I was taught!

and youde certainly come unstuck in energetics if you tried that! heheheh.

I don't see why that's true; surely, if a reaction works with one mole of a substance, it would work in the same proportions with 2 moles.

Furthermore, I have studied chemistry at university. This module, specifically.

YT2095 said in post # :

Im sure there are SOME instances where you could apply that and get away with it, but to treat each molecule as its WHOLE is the way I was taught, same applies to dimetric or trimetric situations, you CANNOT afford to take to the lowest possible int, itll cock up all yer calcs else.

Oh I see, you're totally misunderstanding my point.

This is for reactions like 20C + 20O2 -> 20CO2, which can obviously be cancelled down to C + O2 -> CO2.

I'm not, say, talking about cancelling C6H6 to CH.

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Hmmm... ok, you probably dont realise it but point 2 rellates to point 1.

heres how, in energetics (a euphemism for explosives and the like). it is VITAL to keep ALL in its original form, mole for mole doesnt apply here, as with hydrodynamics just 50% more can double the effect, it can be exponential (and often is).

it does share similarities with your second point that not ALL calcs can or SHOULD be treated like this, thats all

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2 C4H10 + 13 O2 ==> 8 CO2 + 10 H2O

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YT2095 said in post # :

Hmmm... ok, you probably dont realise it but point 2 rellates to point 1.

heres how, in energetics (a euphemism for explosives and the like). it is VITAL to keep ALL in its original form, mole for mole doesnt apply here, as with hydrodynamics just 50% more can double the effect, it can be exponential (and often is).

it does share similarities with your second point that not ALL calcs can or SHOULD be treated like this, thats all

That's energy concerns, which deals with absolute amounts, not comparative calculations.

Furthermore, (and correct me if I'm wrong) I'd think that it's more to do with physical conditions than chemical reactions if the output is increased more than suggested by mass ratio.

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yes and no, mostly yes, but no in the way of dimetric to trimetric conversions with regard to Cyclo compounds, youde expect 1 thrird extra power in yeild, in actual fact for some chems it more than doubles, as I said molarity need not apply here.

it`s to do with bond enrgies and structure stability or the molecules in question, taking them to the lowest possible integers would be not only incorrect procedure for obvious reasons, but more than likely result in utter carnage! to but not too fine a point on it

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C4H9OH + 6O2 --> 4CO2 + 5H2O

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2 C4H10 + 13 O2 ==> 8 CO2 + 10 H2O

Sometimes i really need to read the question.

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WHOA!! please don't get too complex!

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This is correct and balanced:

CH3(CH2)2CH2OH + 6O2 ------------> 4CO2 + 5H2O

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Haha.. I hope you know that I posted this question more than 4 years ago, when I was still in school.

(...and looking back at it now, my typing was absolutely HORRIBLE!)

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butanol as in Butan-1-ol CH3(CH2)2CH2OH ?

Butan-1-ol is C4H9OH

The equation for the complete combustion of Butan-1-ol is...

C4H9OH+6O2 ------> 4CO2+5H2O

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Butan-1-ol is C4H9OH

R Butan-2-ol is C4H9OH

S Butan-2-ol is C4H9OH

and also t-Butanol is C4H9OH

But CH3(CH2)2CH2OH is Butan-1-ol

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