miu,glucoza= 180g/mol
niu=m/miu=630/180mol=3,5mol glucoza
1mol.............................................2mol
C6H12O6---> 2C2H5OH+ 2CO2
3,5mol............................................x
x=niu=7molCO2 Valoare teoretica
Practic, niu= 7x70/100mol=4,9mol
1mol.......1mol
CO2+ Ca(OH)2-->Caco3+ H2O
4,9mol............x
x=niu=4,9molCa(OH)2
c,m= niu/V---> V= 4,9/1 L= =4.9 l