notez a molCH4 , cu M= 16g/mol
b mol C4H10, cu M= 58g/mol
100g amestec...contine.........81,051gC
(16a+58b)g............................(12a+48b)
1200a+4800b=1297a+4700b--->97a=100b---> a/b=100/97
a/b=1/0,97 raport molar, caruia ii corespunde raportul de masa
m,CH4/m,C4H10= 1x16/0,97x58
in (16+56,3) g amestec.............16gCH4..................56,3gC4H10
100g........................................X..................................Y
calculeaza x si y !!!!!!!