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How can we balance oxygen or any other diatomic molecule with a fractional coefficient in a reaction? I mean, if we can balance oxygen atoms in an odd number what happens to other oxygen atom, and can this sort of reaction even happen?

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    $\begingroup$ The use of fractions is irrelevant, as it's only the ratio that matters. If you don't like fractions you can multiply through by the denominator to get all whole numbers. $\endgroup$ – bon Jul 15 '15 at 15:53
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Whenever a chemical reaction is written with fractional coefficients, implicitly it's meant to be interpreted as the amount of substance in moles of each species that is involved. So, it's not really one half of an oxygen molecule reacting, but 0.5 moles of $\ce{O2}$ reacting.

bon's comment is spot-on -- if you are trying to think about a reaction written this way on a molecular level, you need to multiply the reaction through by whatever constant eliminates all the fractions.

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  • $\begingroup$ Brian you mean in molecular level it is impossible to write reaction in Fraction but if the coefficient numbers mean moles we can use Fraction is that true ? and THANK YOU SO SO MUCH and please answer my question in Easy English because i am not native speaker $\endgroup$ – Yasser Jul 15 '15 at 16:06
  • $\begingroup$ @Yasser Yes, that's exactly right. On the molecular level it makes little sense to write in fractions, but on the 'moles level' it's fine. $\endgroup$ – hBy2Py Jul 15 '15 at 16:11
  • $\begingroup$ Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". $\endgroup$ – Martin - マーチン Oct 24 '17 at 8:19
  • $\begingroup$ @Martin-マーチン How would you rewrite? For the life of me I can't concisely frame how to express that the coefficients are to be multiplied into a one-mole quantity of each species. $\endgroup$ – hBy2Py Oct 24 '17 at 14:40
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    $\begingroup$ @hBy2Py IUPAC also recommends enplethy for international usage. I have been trying to use it as of late due to brevity and to sound posh. $\endgroup$ – Linear Christmas Oct 24 '17 at 19:23

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