Un amestec de 1-butena si 3-hexena in raport molar 2:3 se oxideaza cu KMnO4 (H2SO4). Substanta organica formata se dizolva in 1260 g apa formand o sol cu concentratie 37%.Masa de 1 butena din amestec este egala cu ?Raspuns: 140 gEcuatiile reactiilor sunt:

CH2=CH-CH2-CH3 + 5[O] CH3-CH2-COOH + CO2 + H2O

CH3-CH2-CH=CH-CH2-CH3 + 4[O] 2CH3-CH2-COOHSe determina masa de acid propanoic formata

In 100 g solutie sunt 37 g acid propanoic si 63 g H2O

63 g H2O.37 g acid propanoic

1260 g H2O..x

x = 740 g acid propanoic

Se determina numarul de moli de acid propanoic

1 mol CH3-CH2-COOH = 74 g

n = 740/74 = 10 moli CH3-CH2-COOH

Se noteaza cu

a numarul de moli de 1-butena

b numarul de moli de 3-hexena

a/b = 2/3

3a = 2b

b = 1,5aSe determina numarul de moli de CH3-CH2-COOH format din a moli 1-butena si din b moli 3-hexena

1 mol1 mol

CH2=CH-CH2-CH3 + 5[O] CH3-CH2-COOH + CO2 + H2O

a moli..x

x = a moli CH3-CH2-COOH

1 mol.2 moli

CH3-CH2-CH=CH-CH2-CH3 + 4[O] 2CH3-CH2-COOH

b moli.y

y = 2b moli CH3-CH2-COOH

a + 2b = 10

a + 3a = 10

4a = 10

a = 2,5 moli CH3-CH2-COOH

Se determina masa de 1-butena

1 mol C4H8 = 56 gm 1-butena = 2,5 56 = 140 g