Răspuns:
FORMULA CnH2n+2O M= 14n+2+16 g/mol
CnH2n+2O ====> ardere nCO2 +(n+1)H2O
14n+18 g...............................44n
18,4g.....................................35,2g
35,2(14n+18)= 18,4x44n
n=2 -----> C2H6O----alcool etilic
C2H6O + 7/2O2=======> 2CO2+ 3H2O
46g...............3,5mol
18,4g...........x
x=niu=1,4molO2
V,O2=niuxVm=1,4x22,4 l=31,36l
V.aer= 100V,O2/80=156,8 L
Explicație:
