Question

Arene 7.13
Ajutati-ma va rog frumos

Answer 1

m1 C10H8      m2 C7H8
m1 + m2 = 31,2 g      (m1/128 ·10·12 + m2/92 ·7·12)·100/31,2 = 92,3077
(5m1/64 + 7m2/92) ·100 /2,6 = 92,3077
3m1 + 2,9264m2 = 92,3077
0,0735m2 = 1,2923    m2 = 17,5823 g toluen  ( ≈0,2 moli)    m1 = 13,6176 g naftalina  (≈0,1 moli)
a) n HNO3 = 105/100 ·0,3 = 0,315 moli
nHNO3: nH2SO4 : H2O = 1:1:0,5 = 0,315:0,315: 0,1575
m amestec nitrant = 0,315·63 + 0,315·98 + 0,1575·18 = 53,55 g
b) sol. H2SO4 : md = 30,78g  c = 98%   ms = 30,78·100/98 = 31,5g
 mH2O = 31,5- 30,78 = 0,72g
c)  sol. HNO3 : md = 19,845   m H2O = 2,835 - 0,72 = 2,115 g
ms = 21,96g    c = 1984,5/21,96 = 90,368%
d) acidul rezidual contine : 0,15 moliHNO3  (9,45g),   0,315moli H2SO4 (30,78g),  apa : 2,835g + 0,3·18 = 8,235g
ms = 9,45 + 30,78 + 8,235 = 48,465g 
cHNO3 = 945/48,465 = 19,5%
cH2SO4 = 3078/48,465 = 63,5%
e)  Daca la clorurarea fotochimica participa doar toluenul
C6H5-CH3 + 3Cl2 ⇒ C6H5CCl3 + 3HCl
nCl2 = 3n toluen = 0,6 moli  V Cl2 = 0,6··22,4 = 13,44 l