notam hidrocarbura CaHb
notam n = nr. moli C2H6 si m = nr moli CaHb
=> n/m = 2/5 => 2,5n = m
PV = niuRT
niu = m/M.mediu , M.mediu = masa moleculara medie a amestecului de hidrocarburi
p = m/V
=> PM = pRT => M = pRT/P
= 2,68x0,082x(273+91)/2 = 40 g/mol
n + m = n + 2,5n = 3,5n moli amestec hidrocarburi
M.mediu = fractieC2H6xM.C2H6 + fractieCaHbxMCaHb
M.C2H6 = 30 g/mol , M.CaHb = 12a+b g/mol
fractieC2H6 = n/3,5n = 1/3,5 = 0,286
fractieCaHb = 2,5n/3,5n = 0,714
=> 40 = 30x0,286 + 0,714M.CaHb
=> 31,42 = 0,714M.CaHb
=> M.CaHb = 44 g/mol
=> 12a + b = 44
necunoscand nimic despre hidrocarbura, punem conditii pt. numarul de at de C
daca a = 3 => b = 8 => C3H8 => propan
