consideram ca procentele sunt molare = % volumetrice
0,1b 0,1b 0,1b
C3H8 --toC--> C3H6 + H2
1 1 1
0,35b 0,35b 0,35b
C3H8 --toC--> C2H4 + CH4
1 1 1
la C3H8 netransformat avem b-0,1b-0,1b-0,35b-0,35b = 0,1b
0,1b 0,1b
C3H8 ---------> C3H8 netransformat
1 1
------------- ----------------------------
a kmoli C3H8 b kmoli gaze
intrat in reactie rezultate
a)
a = 0,1b+0,35b+0,1b = 0,55b kmoli C3H8 intrat in proces
100m3/22,4 = 4,464 kmoli C3H8
=> 0,55b = 4,464 => b = 8,12
=> C2H4 = 0,35b = 2,84 kmoli => 2,84x22,4 = 63,64 m3 C2H4
b)
%C3H8 transformat in H = 0,1bx100/0,55b = 18,18%
c)
0,35b kmoli n kmoli
C2H4 + 1/2O2 --Ag2O/350oC--> (C2H4)O, etilenoxid (oxid de etena)
1 1
=> n = 0,35b kmoli la 100% ........... la 60%...?
100% ..................... 0,35b kmoli
60% ....................... Cp = 0,21b kmoli
0,21b kmoli md kg
(C2H4)O + H2O --H+--> HO-CH2-CH2-OH
1 62
=> md = 0,21bx62/1 = 13,02b kg obtinut 100%...... la 60%
100% ............... 13,02b kg
60% ................. Cp = 7,812b kg
=> md etandiol = 7,812b = 63,43 kg
c% = mdx100/ms => ms = mdx100/c%
= 63,43x100/95 = 66,77 kg
