100 tone sfecla.......8 tone zahar
2 tone sfecla...........x tone zahar
x=2*8/100=0,16 tone zahar=160 kg zahar
100 kg zahar...........80 kg ramas dupa operatii tehnologice
160 kg...............................x kg ramase
x=160*80/100=128 kg ramase
C12H22O11-zaharoza
C12H22O11 + H2O = C6H12O6(alfa glucoza) + C6H12O6 (beta fructoza)