Question

O metoda petrochimica de sinteza a glicerolului foloseste ca materie prima propena din gazele de cracare. Care este volumul de clor ( masurat in conditii normale ) , de puritate 75%, necesar obtinerii a 690 kg de glicerol, de puritate 80% cu un randament global de 75%?

Answer 1

CH2=CH-CH3+ Cl2---> CH2=CH-CH2Cl+HCl... ................la 500grade
CH2=CH-CH2Cl+  HOH---> CH2=CH-CH2OH
CH2=CH-CH2OH +Cl2---> CH2Cl-CHCl-CH2OH
CH2Cl-CHCl-CH2OH +2HOH---> CH2OH-CHOH-CH2OH + 2HCl
din aceste ecuatii, rezulta ca  1 mol glocerina se obtine utilizand 2molCl2
M(miu),glicerina= 92kg/kmol
m,glicerina pura:  80/100= m,pur/690---> m,pur=552kg
n(niu)= 552kg/92kg/kmol= 6kmolglicerina

mol Cl2= 12 kmol Cl2 pur, la randament 100%
n,Cl2 impur= 12x100/75=>16kmol
n,Cl2 la randament 75%= 16x100/75=>21,33kmol
V,Cl2= 21,33kmolx22,4 mcub/kmol=...................calculeaza !!!!