3H₂C=CH₂ + 2KMnO₄ + 4H₂O ⇒ 3OH-CH₂-CH₂-OH + 2KOH + 2MnO₂
C⁻² -1e⁻ ⇒ C⁻¹ se oxideaza (ag. red) ||3
C⁻² -1e⁻ ⇒ C⁻¹ se oxideaza (ag. red) ||3
Mn⁺⁷ +3e⁻ ⇒ Mn⁺⁴ se reduce (ag.ox) ||2
Cm=n/Vs => n=Cm*Vs = 0,1*0,003 = 0,0003 moli KMnO4
3 moli C2H4.........................2 moli KMnO4
x moli C2H4.........................0,0003 moli KMnO4
x=3*0,0003/2 = 0,00045 moli C2H4
pV=nRT => V=nRT/p = [45*82*283*10^(-8)] /1 = 0,01 L C2H4
