Răspuns:
M,glucoza= 180g/mol
niu,glucoza= m/M=1440/180mol=8mol
1mol.............................................2mol
C6H1206---> 2C2H5OH+2CO2
8mol.....................................x= 16mol, daca randamentul e 100%
practic, se obtin mai putini moli
niu,practic= 16x75/100mol= 12molCO2
1mol.........2mol
CO2+ 2NaOH---> Na2CO3+H2O
12mol........x= 24molNaOH
c,m= niu/V---> V= 24/4 L=6 L
Explicație: