Voi merge pe premisa ca ai vrut sa scrii 224L, si nu 244L
vom porni de la 1 mol de alcan.
Stim ca 1 mol........22,4L (c.n) => pornim de la 22,4L alcan
22,4L 22,4nL 22,4(n+1)L
[tex] C_{n} H_{2n+2} [/tex] + [tex] \frac{3n+1}{2} [/tex]O₂ -> nCO₂ + (n+1)H₂O
22,4L x y
x=22,4nL; y=22,4(n+1)L => x+y=22,4(n+n+1)=22,4(2n+1)L=V final
V final = V initial + 224 = 22,4 + 224 = 246,6 L
V final = 22,4(2n+1)=246,4 => 2n+1=246,4/22,4=11 => 2n = 10 => n=5
Alcanul: C₅H₁₂ - n-Pentanul
