Question

Exercitiul 4,va rog!

Answer 1

4. MCxHy= 1,339*22,4 = 30g/mol

℅C = 80 ; ℅H = 20

nC = ℅C*M/AC*100 = 80*30/1200 = 2 atomi C

nH = ℅H*M/AH*100 = 20*30/100 = 6 atomi H

F.M => C₂H₆ - etan

A) H₃C-CH₃ + Cl₂ (lumină) => H₃C-CH₂Cl + HCl

B) H₃C-CH₃ --> C₂H₄ + H₂

C) C₂H₆ + 7/2 O₂ => 2CO₂ + 3H₂O + Q

a) MC₂H₅Cl = 64,5g/mol

%C = 24/64,5*100 = 37,2
%H = 5/64,5*100= 7,75
%Cl = 35,5/64,5*100 = 55%

c) nt = 10 moli (amestec gaze) 

np = 75*10/100 = 7,5 moli gaze (H2 si C2H4)

pV=nRT => V=[75*82*39*10^(-3)]/ 3,9 = 61,5 L gaze

d) m = 184g amestec

30x+16y = 184 => 46x = 184 => x=4 moli si y=4 moli

1 mol C2H6................3,5*22,4 L O2
4 moli C2H6 ................x=313,6 L O2

CH4 + 2O2 => CO2 + 2H2O + Q

1 mol CH4....................2*22,4L O2
4 moli CH4......................x=179,2 L O2 

VO2 total = 492,8 L

Vaer(20%O2) = 5VO2 = 5*492,8 = 2464 L