1mol n moli 18(n+1) g
CnH2n+2 + (3n+1)/2 O₂ -> nCO₂ + (n+1)H₂O
x moli x*n moli 18x(n+1) g
1mol (n+1) moli 18(n+2) g
Cn+1H2n+4 (3n+4)/2 O₂ -> (n+1)CO₂ + (n+2)H₂O
x moli x(n+1) moli 18x(n+2) g
x(n+n+1)=x(2n+1)=22 moli
x(n+1+n+2)=x(2n+3)=468/18=26moli
facem raport:
(2n+1)/(2n+3)=22/26 => 52n+26=44n+66 => 8n=40 => n=5 deci n+1=6
=> raspuns: hexanul (C₆H₁₄)
