Question

La alchilarea benzenului cu propena se obtine un amestec de reactie care contine izopropilbenzen: diizopropilbenzen: benzen in raport molar 7:2:1. Calculati masa de izopropilbenzen care se obtine, daca se utilizeaza 1950L benzen (ro=0,8g/cm^3)

Answer 1

x moli.....x................x
C₆H₆ + C₃H₆ ⇒ C₆H₅-C₃H₇
y moli......2y.................y
C₆H₆ + 2C₃H₆  ⇒ C₆H₄-(C₃H₇)₂
z moli.......z
C₆H₆ ⇒ C₆H₆ (nereactionat)

ρ = m/V ⇒ m = ρ×V = 1950000cm³×0,8g/cm³ = 1560000g = 1560kg C₆H₆

n C₆H₆ = 1560/78 = 20 kmoli

x/y = 7/2 ⇒ y = 2x/7 ⇒ y = 4 kmoli izopropilbenzen
x/z = 7/1 ⇒ z = x/7 ⇒ z = 2 kmoli benzen nereactionat
 
x+y+z = 20 ⇔ x+2x/7+x/7 = 20 ⇔ 10x/7 = 20 ⇔ 10x = 140 ⇒ x = 14 kmoli

MC₆H₅-C₃H₇ = 120kg/kmol

n = m/M ⇒ m = n×M = 4×120 = 480 kg izopropilbenzen