Question

Un amestec de naftalina si benzen contine 8,96% H. Care este compozitia masica a amestecului de hidrocarburi?
Help please

Answer 1

notam cu

a = nr. moli naftalina C8H10

b = nr moli benzen C6H6

1 mol C10H8 ......... 128 g ............... 8 g H

a moli ................. 128a g ............... 8a g H

1 mol C6H6 ......... 78 g ............... 6 g H

b moli .................  78b g ............... 6b g H

=>

       100% amestec .................................... 8,96% H

(128a+78b) g amestec .............................. (8a+6b) g H

=> 11,16(8a+6b) = 128a + 78b

=> 38,71a = 11,04b => 3,5a = b

raportul C10H8 : C6H6 = a : b = a : 3,5a /:a = 1 : 3,5

=> ca avem 1 mol C10H8 + 3,5 moli C6H6

=> 128a = 128 g C10H8

=> 78b = 273 g C6H6

=> amestec = 401 g

128x100/401 = 31,92% C10H8

100% - 31,92 = 68,08% C6H6