Question

Ce volum de etena la 10*C si o atmosfera poate fi oxidat la glicol cu 3 ml solutile KMnO4 0,1M?

Answer 1

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