4.a) C6H12O6 --> 2CH3-CH2-OH + 2CO2
b) MC6H12O6=180g/mol
n=m/M=720/180=4 moli glucoza
1mol C6H12O6..............2mol CH3-CH2-OH
4mol...............................x=8mol
MCH3-CH2-OH=46g/mol
n=md/M==>md=n*M=8*46=368g etanol
Cp=md/ms*100==>ms=md*100/Cp=368*100/10=3680g
(C6H10O5)n + nH2O --> 4C6H12O6 ===>162n=720==>n=4.44 moli
n=m/M==>m=n*M=4.44*162=719.199g .. nu sunt sigur aici..
