notam a moli de CH4, miu = 16 g/mol
notam b moli de C2H6, miu = 30 g/mol
16 g CH4 ......................... 4 g H
16a g CH4 .................... 4a g H
30 g C2H6 ...................... 6 g H
30b g C2H6 ..................... 6b H
16a+30b .......................... 100%
4a+6b ............................. 22,58%
4,43(4a+6b) = 16a+30b
=> 1,71a = 3,42b
=> a = 2b
=> raportul CH4 : C2H6 = a : b = 2b : b /b
=> raportul CH4 : C2H6 = 2 : 1 => E.
