Can we use fractional coefficients in stoichiometric equations?

So is #H_2(g) + 1/2O_2(g) rarr H_2O(l)# right?

Now I cannot have HALF a dioxygen molecule, but I certainly can have half a mole, i.e. a #16*g# mass of dioxygen. And this combines with a mole of dihydrogen, i.e. a #2*g# mass of #H_2#, to give a molar quantity of water, i.e. an #18*g# mass of water.

The stoichiometric coefficients used in the given equation are thus no more than tools. They may be appropriate or inappropriate given a specific scenario. When we assess stoichiometric equivalence, I maintain that a half-integral coefficient can certainly be useful, and correct, and therefore right. And, for me at least, when we write out the alternative.....

#2H_2(g) + O_2(g) rarr 2H_2O(l)#

....when we calculate equivalent masses, there is the possibility of ballsing something up. Do I divide by two, or do I multiply by two...to find mass equivalence? And still, after many years, it is NOT immediately obvious to me.....on the other hand...for...

#H_2(g) + 1/2O_2(g) rarr H_2O(l)#

..the mass equivalence is directly obvious to me.

We know that all chemical equations observe stoichiometry.

And the answer to your question is that by the criteria I have advanced.....#H_2(g) + 1/2O_2(g) rarr H_2O(l)# IS RIGHT! They are a means to an end to represent stoichiometry, i.e. conservation of mass and charge. The equation is there to serve me, to help me in calculating stoichiometric equivalence. I am not here to serve some abstract notion of chemical equations.

Likewise.....#HC-=CH(g) +5/2O_2(g)rarr 2CO_2(g) +H_2O(l)# is also correct, as is...

#C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(l)# because the representation absolutely CONSERVES MASS and CHARGE. Happy? Not everyone will be.