Question

.Prin arderea unui amestec echimolecular a doua hidrocarburi saturate
omoloage se obtin 968 g CO2 si 468 g apa. Care sunt cele doua
hidrocarburi ?

Answer 1

CnH2n+2 +(3n+1)/2 O2 -> nCO2 + (n+1)H2O
   1 mol ............................ n moli....... (n+1) moli
   x moli ......................... x*n moli..... x(n+1) moli  
CmH2m+2 +(3m+1)/2 O2 -> mCO2  +  (m+1)H2O
    1 mol .............................m moli ..... (m+1) moli
    x moli ...........................x*m moli..... x(m+1) moli
r.molar = x:x
MCO2=12+2*16=44
MH2O=2*1+16=18
968:44=22 moli CO2
468:18=26 moli H2O
n*x+m*x=22
(n+1)x+(m+1)x=26 <=> n*x+m*x+2x=26 <=> 22+2x=26 =>x=2 
(n+m)*2=22 => n+m =11
Cele 2 hidrocarburi sunt omoloage => n=5 ; m=6
C5H12 pentan ; C6H14 hexan