Question

prin clorurarea fotochimica a CH4 se obtine un amestec de CH3Cl si CH2Cl2 care contine 80,5% Cl. Determinati:
a) Raportul molar CH3Cl : CH2Cl2
b) Volumul de Cl necesar pentru a se obtine 441 kg amestec
c) Masa amestecului reactant (masa de CH4 + cea de Cl2) care contine 9,033 x 10 la puterea 23 atomi de Cl, stiind ca reactantii se consuma integral.

Answer 1

a) MCH3Cl=50,5g/mol

MCH2Cl2=85g/mol

MCH4=16g/mol

50,5x+85y.........................................35,5x+71y
100....................................................80,5

4065,25x+6842,5y=3550x+7100y => 515,25x=257,5 => x/y=1/2

CH3Cl:CH2Cl2=1:2

b) 50,5x+2*85y=441=>220,5x=441 =>x=2 moli

x/y=1/2=>y=2x=>y=4 moli

1 kmol Cl2.......................1 kmol CH3Cl
2 kmoli Cl2.....................2 kmoli CH3Cl

2 kmoli Cl2.............................1 kmol CH2Cl2
x=8 kmoli...............................4 kmoli CH2Cl2

nr total kmoli Cl2=10 kmoli => V=224m3 Cl2

Answer 2

n1 kmoli  CH4 + Cl2 ⇒ CH3Cl + HCl
n2 kmoli  CH4 + 2Cl2 ⇒ CH2Cl2 + 2HCl
35,5(n1 + 2n2)·100/(50,5n1 + 85n2) = 80,5
3550n1 + 7100n2 = 4065,25n1+ 6842,5n2
257,5 n2 = 515,25n1   n2 = 2n1         n1/n2 = 1/2
b) 50,5n1 + 85·2n1 = 441    220,5n1= 441  n1 = 2 kmoli     n2 = 4kmoli
 nCl2 = n1 + 2n2 = 10kmoli   V = 224m³
c) amestec reactant : (n+ 2n) moliCH4  + (n + 2·2n)moli Cl2
m amestec = 3n·16 + 5n·71= 403n g
nCl2 = 5n         nCl = 10n = 9,033·10²³ /6,023ₓ10²³    10n = 1,5  n = 0,15moli
m amestec = 403ₓ0,15 = 604,5g