Se alchileaza benzenul cu etena pentru obtinerea etilbenzenului.In vasul de alchilare se obtine o masa de reactie de compozitia: 57%C6H6, 35%etilbenzen, 6% dietilbenzen, si 2% trietilbenzen.Daca in procesul de fabricatie se obtin 184 t masa de reactie, sa se determine:a) Ce cantitate de benzen s-a introdus in procesul de fabricatie?Din benzenul introdus in fabricatie cat s-a transformat in etilbenzen,dietilbenzen, trietilbenzen si cat a ramas netransformat? (in% masa)c)In ce raport molar s-au introdus in procesul de fabricatie benzenul si etena?

se determina masa din fiecare component din amestecul final:100 kg amestec…..57 kg benzene…..35 kg etilbenzen…….6 kg dietilbenzen…2 kg trietilbenzen184000 kg amestec….x………………y……………………z…………………………tx=184000∙57/100=104880 kg benzeny=184000∙35/100=64400 kg etilbenzenz=184000∙6/100=11040 kg dietilbenzent=184000∙2/100=3680 kg trietilbenzen1 kmol C6H6 = 78 kg1 kmol C6H5-CH2-CH3 = 77 + 29= 106 kg1 kmol C6H4(CH2-CH3)2 = 76 + 58=134 kg1 kmol C6H3(CH2-CH3)3 = 75+87=162 kg

Ecuatiile reactiilor chimice:C6H6 + C2H4 → C6H5-CH2-CH3

C6H6 + 2C2H4 → C6H4(CH2-CH3)2

C6H6 + 3C2H4 → C6H3(CH2-CH3)3

Se calculeaza numarul de moli de benzene si numarul de kmoli de etena din care se obtin substantele din amestecul final

1 kmol……1 kmol…..106 kgC6H6 + C2H4 → C6H5-CH2-CH3

x1 kmoli…..y1 kmoli….64400kgx1=64400/106=607,54 kmoli C6H6

y1=64400/106=607,54 kmoli C2H4

1 kmol….2 kmol…..134 kgC6H6 + 2C2H4 → C6H4(CH2-CH3)2

x2 kmoli…y2 kmoli…11040kgx2=11040/134=82,38 kmoli C6H6

y2=11040∙2/134=164,77 kmoli C2H4

1 kmol….3 kmol…..162 kgC6H6 + 3C2H4 → C6H3(CH2-CH3)3

x3 kmoli…y3 kmoli…3680kgx3=3680/162=22,71 kmoli C6H6

y3=3680∙3/162=68,14 kmoli C2H4

Numarul de kmoli de C6H6 netransformati=104880/78=1344,61 kmoli C6H6

Numarul de kmoli de C6H6 intrati in reactie: 607,54+82,38+22,71+1344,61=2057,24 kmoli C6H6

Numarul de kmoli de C2H4 intrati in reactie: 607,54+164,77+68,14=840,45 kmoli C2H4

masa totala de C6H6 = 2057,24∙78=160464,72 kg C6H6 = 160,46 t C6H6

masa de C6H6 netransformat = 104880 kg C6H6 =104,88 t C6H6

masa de C6H6 transformat in etilbenzen= 607,54 ∙78=47388,12 kg C6H6=47,38 t C6H6

masa de C6H6 transformat in dietilbenzen = 83,38∙78=6503,64 kg C6H6=6,50 t C6H6

masa de C6H6 transformat in trietilbenzen = 22,71∙78=1771,38 kg C6H6 =1,77 t C6H6

Procentele de C6H6 transformat in diversii component:160,46 t C6H6 introdus…104,88t C6H6 netransformat…..47,38t C6H6 transformat in etilbenzen100 t ………………………….x…………………………….yx=100∙104,88/160,46=65,36% C6H6 netransformaty=100∙47,38/160,46=29,52% C6H6 transformat in etilbenzen

160,46 t C6H6 introdus…….6,50t C6H6 transformat in dietilbenzen …..1,77t C6H6 transformat in trietilbenzen100 t ………………………….z………………………………………….tz=100∙6,5/160,46=4,05 % C6H6 transformat in dietilbenzent=100∙1,77/160,46=1,1 % C6H6 transformat in trietilbenzen

b) raportul dintre benzen si etena=2057,24/840,45=2,44