Question

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 e.....

Answer 1

Mbutena=56g/mol si Mhexena=84g/mol
raport de 2:3 => 2x moli butena si 3x moli hexena
n=m/M => m=n*M
mbutena=56*2x=112x (vom folosi mai incolo pentru determinarea masei)
1mol...........................................................74g
CH2=CH-CH2-CH3 + 5[O] -> CO2 + H2O + CH3-CH2-COOH
2x..............................................................148x
1mol.......................................................148g
CH3-CH2-CH=CH-CH2-CH3 + 4[O] ->2CH3-CH2-COOH
3x...........................................................444x
macid=592x g
c/100=macid/(macid+mapa) => 37/100=592x/(592x+1260)
59200x=21904x+46620 => 37296x=46620 => x=1,25moli
mbutena=112*1,25=140g