Un amestec de metan si etan are un continut procentual in carbon 78,75%.
Rezolvare
notam cu a = nr. moli CH4 si b moli C2H6
M.CH4 = 16 g/mol, M.C2H6 = 30 g/mol
16g CH4 .................... 12g C
16a g CH4 ................. 12a g C
30g C2H6 ................. 24g C
30b g C2H6 .............. 24b g C
=>
100% amestec ............................ (16a+30b) g amestec
78,75% C .................................... (12a+24b) g C
=> 15,238a + 30,476b = 16a + 30b
=> 0,762a = 0,476b => b = 1,6a
Volumul de CH4 = 22,4a L
Volumul de C2H6 = 22,4b = 35,84a L
=> %CH4 = 22,4a100/35,84a = 62,5%
=> %C2H6 = 100-62,5 = 37,5%
=> B) aproximativ
