C₄H₁₀ -> CH₄ + C₃H₆
C₄H₁₀ -> C₂H₆ + C₂H₄
Cu HCl pot fi tratate doar alchenele, deci alcanii vor iesi din amestec
CH₂=CH₂ + HCl -> CH₃-CH₂-Cl
CH₂=CH-CH₃ + HCl -> CH₃-CH(Cl)-CH₃ (paranteza inseamna ramificatie)
Notam raportul molar etena:propena x:y
reamintesc faptul ca m=n×M (masa de substanta sau amestec=numar de moli × masa molara)
1mol 64,5g
CH₂=CH₂ + HCl -> CH₃-CH₂-Cl
xmoli 64,5x g
1mol 78,5g
CH₂=CH-CH₃ + HCl -> CH₃-CH(Cl)-CH₃
ymoli 78,5y g
64,5x+78,5y................100%
35,5(x+y)...................49,7%
3550x+3550y=3205,65x+3901,45y
344,55x=351,45y => x=y (aproximativ)
deoarece reactiile decurg mol:mol => raportul etena:propena=1:1
