Question

Se obtine zaharoza prin deshidratarea intrermoleculara a54g glucoza si 36g fructoza la un randament al reactiei de 75%
a) Ce cantitate de zaharoza se obtine
b) Cati moli de sulfat acid de metil pot eterifica zaharoza formata

Answer 1

m glucoza = 54 g
m fructoza = 36 g
M glucoza/fructoza = 180.16 g/mol
n glucoza = 0.3 mol glucoza
n fructoza = 0.2 mol fructoza (reactant limitativ)

raportul molar fructoza:glucoza = 1:1 asadar:

a.
glucoza + fructoza = zaharoza
1 mol fructoza...1 mol zaharoza
atunci 
0.2 mol fructoza ... 0.2 mol zaharoza (cantitatea teoretica)
randament = 75%
deci: 
randament = (cant. practica / cant. teoretica) x 100
75 = (cant. practica/0.2) x 100
cant. practica = (75 x 0.2) /100
cant. practica = 0.15 mol
masa zaharoza = 0.15 mol x M Zaharoza = 0.15 x 342.3 = 51.35 g zaharoza
M zaharoza = 342.3 g/mol

b. daca se eterifica toate cele 8 grupari hidroxil atunci:
 Zaharoza + 8 CH3-OSO3H <=> eterul corespunzator + 8 H2SO4
1 mol zaharoza consuma 8 mol CH3-OSO3H
atunci
0.15 mol zaharoza consuma y mol CH3-OSO3H
y=(0.15 x 8 ) / 1 = 1.2 mol CH3-OSO3H