Question

Se deshidrateaza 171 g amestec de propanol si butanol si se obtin 56 L de gaze. Compozitia in procente molare a amestecului de alcooli?

Answer 1

MCH3-CH(OH)-CH3 =60g/mol
MC4H10O=74g/mol

CH3-CH(OH)-CH3 ---> CH2=CH-CH3

C4H8O ---> CH3-CH=CH-CH3

60x+74y=171        ||==>  60x+74y=171
x+y=56/22.4=2.5  ||(-60) -60x-60y=-150
                                          /      14y = 21 => y=1.5 moli

60x+74*1.5=171==>60x+111=171==>60x=60 =>x=1mol
nr moli total = 1.5+1=2.5 moli

%C3H8=1.5/2.5*100=60%

%C4H10=1/2.5*100=40%