Question

Se oxideaza 74 kg de 1 butanol cu KMnO4 si H2SO4.Acidul carboxilic obtinut se dizolva ,dupa separare ,in 80 kg de apa.Sa se calculeze concentratia procentuala a solutiei obtinute ,precum si volumul solutiei de KMnO4 de concentratie 0.4M care se consuma in reactia de oxidare.
A.51.5% ,8 dm^3 solutie de KMnO4 0.4M
B.52.38%,2000 dm^3 solutie de KMnO4 0.4M
C.48%,2 dm^3 solutie de KMnO4 0.4M
D.52.38%,2000 l solutie de KMnO4 0.4 M

Answer 1

CH2(OH)-CH2-CH2-CH3
n=m/M=74/74=1 kmol alcool
5CH2(OH)-CH2-CH2-CH3 + 4KMnO4 + 5H2SO4 ---> 5HOOC-CH2-CH2-CH3 + K2SO4 + 4MnSO4 + 10H2O
C(-1) -4e ---> C(+3) se oxidează (ag. red) |||5
Mn (+7) +5e >>> Mn (+2) se reduce (ag.ox) |||4
CH3-CH2-CH2-COOH + H2O---> CH3-CH2-CH2-COO + H3O
ms=md+mH2O=74+80=154kg
md=74kg
Cp=md×100/ms=7400/154=48%
5kmoli alcool.......4kmoli KMnO4
1kmol alcool.......x=0.8 kmoli
Cm=n/Vs==>Vs=n/Cm=0.8/0.4=2m3=2000dm3
MC4H10O=74kg/kmol