Question

Stie cineva ? M ar interesa mai mult raspunsul de la VI si VII ...deeeeci?

Answer 1

n1 moli  CH4 + Cl2 ⇒CH3Cl + HCl
 n2 moli CH4 + 2Cl2 ⇒CH2Cl2 + 2HCl
n3 moli CH4 + 3Cl2 ⇒ CHCl3 + 3HCl
n4 moliCH4 nereactionat
n1/2 = n2/2 = n3/1 = n4/1 = n / 6      n1= n2 = n/3    n3 = n4 = n/6
n1+ 2n2 + 3n3 = 15,68/22,4 = 0,7 moli
2. n/3 + 2n/3 + n/2 = 0,7   1,5n = 0,7    n = 7/1,5 = 14/3 moli CH4
VCH4 pur = 14/3 ·22,41 = 104,58 l
V cu imp. = 104,58·100/80 = 130,725 l
3. agent frigorific
VII. CnH(2n+1)Br      80·100/(14n + 81) = 65      14n + 81 = 123     14n = 42    n= 3    (C3H8)
Br-CH2       -     CH2     -      CH3           CH3       -        CH2Br     -            CH3
      ↓                   ↓                   ↓                 ↓                      ↓                            ↓
C-primar   C-secundar    C-primar     C-primar        C- secundar    C- primar
3. 1,1 dibromopropan ; 1,2 dibromopropan ; 1,3 dibromo propan;
2,2 dibromopropan
VIII. butan  (AlCl3; 50-100°C) ⇄ izobutan
p.f butan> p.f izobutan
3. imbunatatirea calitatilor benzinelor