Question

se supun mononitrarii 880kg amestec ce contine bensen si naftalina, in raport molar de 4:1. Calculati masa solutiei de HNO3, de concentratie 63%, exprimata in kg, necesara amestecului nitrat, utilizat in reactia de nitrare

Answer 1

x..............x....................x..............x
C₆H₆ + HO-NO₂ ⇒ C₆H₅-NO₂ + H₂O
y.................y...................y..............y
C₁₀H₈ + HO-NO₂ ⇒C₁₀H₇-NO₂ + H₂O

78x+128y = 880 ⇔ 440y = 880 ⇒ y = 2 kmoli C₁₀H₈
x/y = 4/1 ⇒ x = 4y ⇒ x = 8 kmoli C₆H₆

n amestec = n HNO₃ = 10 kmoli

md HNO₃ = 10*63 = 630kg 

C = md/ms*100 => ms = 63000/63 = 1000kg sol. HNO₃