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
