Question

145 g amestec format din propena si 1-butena aflate in raport molar de 1:2 se hidrogeneaza,formand 120 g amestec final de alcani.Stiind ca randamentul hidrogenarii 1-butenei a fost de 75%,sa se determine numarul total de moli de alcani in amestecul final.

Answer 1

MC3H6=42g/mol
MC4H8=56g/mol

42x+56y=145  ==> 42x+56*2x=145 => 154x=145 =>x=0.941558 moli C3H6

x/y=1/2==>y=2x=>y=2*0.941558=1.88*0.75=1.4123 moli C4H8

CH2=CH-CH3 + H2 ---> CH3-CH2-CH3

CH3-CH=CH-CH3 + H2 ---> CH3-CH2-CH2-CH3 

nC3H6=nC3H8=1.88 => mC3H8=0.941558*44=41.428552g C3H8

nC4H8=nC4H10=0.7 => mC4H10=1.4123*58=81.9134g C4H10


nr total moli = 0.941558 +1.4123 =2.353 moli amestec alcani