Question

Care este Volumul solutiei acide de KMnO4 , 1M, care oxideaza 1L solutie de etanol de concentratie 46% si densitatea de 1,3 g/cm cubi

Answer 1

ro=ms/Vs=> ms=ro*Vs = 1,3*1000 = 1300g sol alcool

C=md/ms*100 = > md=C*ms/100 = 46*1300/100 = 598g alcool

n=m/M=598/46 = 13 moli C2H6O

5CH₃-CH₂-OH + 4KMnO4 + 6H2SO4 -> 5CH3-COOH + 2K2SO4 + 4MnSO4 + 11H2O

C⁻¹ -4e⁻ ⇒ C⁺³ se oxideaza (ag.red)        ||5
Mn⁺⁷ +5e⁻ ⇒ Mn⁺² se reduce (ag.ox)      ||4

5 moli alcool................4 moli KMnO4
13 moli alcool.............x=10,4 moli KMnO4

Cm=n/Vs => Vs=n/Cm = 10,4/1 = 10,4 L sol KMnO4

Verifica !!!