Question

Cel care rezolva corect si complet, cu explicatii, primeste COROANA si o recomandare de catre mine divinitatilor sa-l accepte in heaven.​

Answer 1

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