Question

Ce volum de etena , la 10 grade si o atmosfera poate fi oxidat la glicol cu 3 ml soluie KMnO4 0,1N?

Answer 1

5CH2=CH2 + 2KMnO4 + 3H2SO4 => 5CH2(OH)-CH2(OH) + K2SO4 + 2MnSO4 + 3H2O

C(-2) --> C(-1) ||5

C(-2) --> C(-1) ||5

Mn(+7) --> Mn(+2) ||2

Cn=md/Eg*Vs=>md=0,1*31,6*0,003=0,00948g KMnO4

5 moli C2H4...........2*158g KMnO4
x moli C2H4...........0,00948g KMnO4

x=0,00015 moli C2H4

pV=nRT=>V=nRT/p=0,00015*0,082*283/1=0,0034809 L = 3,481cm3 C2H4