Question

Cine stie cum se rezolva?

Answer 1

a) p/100=Vpur/Vimpur => Vpur=p*Vimpur/100=89,6*1250/100=1120 m³ acetilena pura
22,4m³..............1kmol
C2H2 + HCl -> C2H3Cl
1120m³............x=50kmoli
90/100=npractici/50 => npractici=45 kmoli
stim ca si din procesul etenei, vom obtine tot 45 kmoli de CV
deoarece stim randamentele reactiilor, calculam randamentul procesului (rp)
rg=r2*r3*100/10000=0,8*0,75*100=60%
28kg.................1kmol
CH2=CH2 -> CH2=CH-Cl (ecuatia procesului)
y=1260kg.......45kmoli
60/100=1260/mteoretica => mteoretica=2100kg etena=2,1t etena
b)
nkmoli...................62,5n kg
nCH2=CH-Cl -> (-CH2-CH(Cl)-)n
45kmoli................2812,5kg
95/100=mpractica/2812,5 => mpractica=2812,5*95/100=2671,875kg
94/100=2671,875/mimpura => mimpura=2671,875*100/94=2842,42kg PVC impur