(C6H10O5)n + H2O => C6H12O6
C6H12O6 => 2C2H5-OH + 2CO2
md alcool = 90*1840/100 = 1656kg
practic: 1656kg alcool
teoretic: 165600/75 = 2208kg alcool
n alcool = 2208/46 = 48 kmoli
n alcool = n C6H12O6 = n (C6H10O5)n = 48 kmoli
a) m amidon = 162*48 = 7776 kg
m amidon pur = 0,9*7776 = 6998,4kg
mp = 80*6998,4/100= 5598,72 kg amidon
b) 25%........................5598,72 kg
100%............................x = 22394,88kg
m cartofi = 22394,88-5598,72 = 16796,16 kg
c) m glucoza pura = 48*180 = 8640kg
m impura = 864000/90 = 9600kg C6H12O6
Nu sunt sigur de logica si rezultate.. Daca ai avea niste rezultate/barem ar fi de folos :)