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
