Răspuns :
Din amestec se hidrogeneaza butadiena
C4H6 + 2 H2 => C4H10
54 g butadiena...........2 * 22,4 l H2
x g butadiena............3,54 l H2
x = 4.3 g
masa butan = 5 - 4.3 = 0.7 g butan
% butan transformat = masa butadiena/masa amestec * 100
% = 4,3/5 * 100 = 86%
C4H6 + 2 H2 => C4H10
54 g butadiena...........2 * 22,4 l H2
x g butadiena............3,54 l H2
x = 4.3 g
masa butan = 5 - 4.3 = 0.7 g butan
% butan transformat = masa butadiena/masa amestec * 100
% = 4,3/5 * 100 = 86%
CH3-CH2-CH2-CH3 ===> CH2=CH-CH=CH2
Doar butadiena este supusa hidrogenarii..
54g,,,,,,,,,,,,,,,,,,,,,,,,,,,..44,8L
CH2=CH-CH=CH2 + 2H2 ---> CH3-CH2-CH2-CH3
x=4,2669g.................3,54L
Prin diferenta aflam masa butanului netransformat:
5-4,2669=0,733g C4H10
%C4H10 transformat=% C4H6 transformata = 4,2669/5*100=85,338%