Question

raportul molar KMnO4/H2SO4 respectiv KMnO4/H2O la oxidarea a 0,6 moli de 2 metil 1,4 hexadiena este

Answer 1

1mol..............................................8moli
CH2=C(CH3)-CH2-CH=CH-CH3 + 8[O] -> CO2 + H2O + CH3-CO-CH2-COOH + CH3-COOH
0,6moli........................................4,8moli
2moli..........3moli.......5moli
2KMnO4 + 3H2SO4 -> 5[O] + K2SO4 + 2MnSO4 + 3H2O
1,92moli....2,88moli..4,8moli
nKMnO4:nH2SO4=1,92:2,88=2:3
1mol..............................................4moli
CH2=C(CH3)-CH2-CH=CH-CH3 + 4[O] -> HO-CH2-COH(CH3)-CH2-CH(OH)-CH(OH)-CH3
0,6moli........................................2,4moli
2moli.........1mol....3moli
2KMnO4 + H2O -> 3[O] + 2KOH + 2MnO2
1,6moli.....0,8mol..2,4moli
nKMnO4:nH2O=1,6:0,8=2:1