Question

raportul atomic C:H intr-un amestec gazos care contine etan si o alchena A este 3 :8
a . Determinati formula moleculara a alchenei A stiind ca amestecul de hidrocarburi contine 75 % etan in procente molare .

Answer 1

etan = C2H6 , avem a moli , M= 30 g/mol

alchena = CnH2n , avem b moli , M= 14n g/mol

a + b ............................ 100% amestec

a .................................. 75%

a+b = 1,333a => b = 0,333a

(2a+nb)/(6a+2nb) = 3/8

=> 18a + 6nb = 16a + 8nb => 2a = 2nb => a = nb

n = a/b = a/0,333a = n = 3 => C3H6, propena