Question

Problema 3 roman , vă rog

Answer 1

a. Calcularea alcoolului etilic pur: 1*94/100= 0.94 l
Calcularea masei alcoolului:
0.76 g........1 ml
x g..............940 ml (0.94 l)
x= 714.4 g

46*2 g etanol..........54 g butadiena
714.4...........................y
y= 419.32

Aplicarea randamentului: 419.32*80/100= 335.45 g butadiena

b. CH2=CH-CH=CH2 ===> (CH2-CH=CH-CH2)

Toata masa de monomer se transforma in polimer (la randament de 100% masa monomer=masa polimer)

Aplicarea randamentului: 335.45*90/100=309.91 g polimer