Question

prin clorurare metan la lumina se obtine un amestec format din monocloroetan si dicloroetan ce contine 80,5% clor.
A) raport molar ch3cl: ch2cl2
B) volumul cl necesar obtinerii a 441 kg de amestec reactie
C) masa amestec de reactie ce contine 9,033*10la puterea 23 atomi clor.

Answer 1

 a)

(1)  CH4 + Cl2 -> CH3Cl + HCl
(2)  CH4 + 2Cl -> CH2Cl2 + 2HCl

x moli CH3Cl si  y moli CH2Cl2

M CH3Cl=50,5 și  M CH2Cl2=85

mCH3Cl=50,5x

mCH2Cl2=85y

50,5 x g CH3Cl........35,5x g Cl   si in  85y g CH2Cl2......71y g Cl 

== > m Cl= 35,5x+71y

(50,5x+85y) g amestec.......(35,5x+71y) g Cl

100 g amestec......................80,5 g Cl

==>y=2x  sau x/y=1/2

b)din ec. reactiilor :

(1) la un mol Cl2.......1mol CH3Cl    

(2)   2moli Cl2.......1 mol CH2Cl2

         4 moli Cl2......2 moli CH2Cl2

==> pt. un amestec de 3 moli [ 1*50,5 g+ 2*85g] sunt necesari 5 moli Cl2

220,5 g amestec ..........5*22,4 l Cl2

441000g........................w  =224000 l (224mc)

c) 

100g amestec......80,5 g Cl =2,26 moli Cl= 2,26*6,033* 10²³= 13,6*10²³ at. Cl

     z g amestec.........................................................................9,033*10²³ at Cl

z= 66,41 g amestec