Calculam masa de alcool din solutie
92/100=md/300 => md=276g etanol
notam cu a, nr de moli de alcool ars
determinam masa de apa rezultata la ardere, ce ramane in solutie
1mol.................................3*18g
C2H6O + 3O2 -> 2CO2 + 3H2O
a moli................................54a g
m=n*M=a*46=46a g alcool ars
malcoolramasa=mdfinal=mdf=276-46a
msf=ms-46a+54a=ms+8a=300+8a
27,71/100=(276-46a)/(300+8a)
276-46a=83,13+2,2168a
48,2168a=192,87
a=4moli
malcoolars=46a=184g
%alcoolars=184*100/276=66,66%