Question

prin arderea a 420 g amestec de etan si pentan se obtin 649,6 l CO2.Determinati  compozitia  procentuala molara si masica a amestecului de hidrocarburi.

Answer 1

n1 moli C2H6 + 3,5 O2 ⇒ 2CO2 + 3H2O
n2  moli C5H12 + 8O2 ⇒ 5CO2 + 6H2O
30n1 + 72n2 = 420                                   n1+ 2,4n2 = 14
nCO2 = 2n1 + 5n2 = 649,6/22,4 = 29     2n1 + 5n2 = 29
                                                                ------------------------
                                                                    0,2n2 = 1   n2 = 5moli pentan ;
n1 = 2moli etan
%etan = 2·100/7 = 28,57%       % C2H6 = 60·100/420 = 14,285%
% pentan = 5·100/7 = 71,428%    % C5H12 = 5·72·100/420 = 85,714%