Question

O proba de faina , ce contine 64,8% amidon , este supusa hidrolizei enzimatice . Stiind ca s-au obtinut 18 g de glucoza , determinati masa probei de faina supusa hidrolizei enzimatice , exprimata in grame . Stie cineva cum se rezolva aceasta problema ?

Answer 1

162n g................................180n
(C6H10O5)n + nH2O = nC6H12O6
x............................................18g

M C6H12O6 = 180 g/mol
M (C6H10O5)n = 162 n g/mol

x= 18×162/180= 16.2 g amidon pur = m pur
puritatea = m pur×100/m impur
m impur = 16.2 × 100/ 64.8 = 25g proba faina