OH
|
3H₂C=CH-CH₃ + 2KMnO₄ + 4H₂O => 3H₂C-CH-CH₃+ 2KOH + 2MnO₂
|
OH
C⁻² -1e⁻⇒ C⁻¹ se oxideaza (ag.red) ||3
C⁻¹ -1e⁻⇒ C⁰ se oxideaza (ag.red) ||3
Mn⁺⁷ +3e⁻ ⇒ Mn⁺⁴ se reduce (ag.xo) ||2
3 moli C3H6................2 moli KMnO4
2 moli C3H6................x= 1,333 moli KMnO4
Cm=n/Vs => Vs=1,333/0,1 = 13,33 L KMnO4
