3,12g compus.........3,36l CO2...........3,24gH2O
104g(1mol)......................x..........................y
x=112lCO2
y=108gH2O
n= V/Vm; n,CO2= 112l/22,4l/mol=5molCO2------> 5molC
n/m/M; n,H2O= 108g/18g/mol= 6molH2O-----> 12mol H
mC+mH= 5X12+12X1=72g pentru a obtine M,masa molara, e nevoie de
104-72=>32g.. adica ------> 2 molO
C5H12O2
