Question

In urma clorurarii metanului se obtin CH3Cl si CH2Cl2 in raport molar 1:2
a) ce volum de metan cantitativ se consuma astfe incat amestecul rezultat sa contina 8,5g CH2Cl2?
b) afla masa de clor( exces 20 %) necesara obtinerii amestecului considerat
c) ce cantitate de solutie HCl de concentratie 36,5% se obtine in urma clorurarii ?

Answer 1

x............x............x...........x
CH4+Cl2-->CH3Cl+HCl

2x..........4x..........2x.............4x
CH4+2Cl2-->CH2Cl2+2HCl

a) n=m/M=8,5/85=0,1 moli CH2Cl2
2x=0,1=>X=0,1/2=0,05 moli
n=V/Vm=>V=n×Vm
V=3X×22,4=3×0.05×22,4=3,375 l CH4 consumat
b) n=m/M=>m=n×M
m=5X×71=5×0,05×71=17,75 g Cl2 consumat
%consumat=100-20=80%
consumat=m total×80/100
m total=consumat×100/80
m total=17,75×100/80=22,18 g Cl2 introdus
c) n=m/M=>m=n×M
m=5X×36,5=5×0,05×36,5=9,125 g HCl
c=md/ms×100=> ms=md×100/c
ms=9,125×100/36,5=25 g soluție HCl