Question

Un amestec de 23,4 grame alcool etilic si fenol reactioneaza cu 6,9 g sodiu. Stabiliti volumul solutiei de NaOH de concentratie 0,25 M care reactioneaza cu acest amestec. R: 800 ml

Answer 1

C6H5-OH + Na = C6H5-ONa + 1/2 H2
C2H5OH + Na = C2H6-ONa +  1/2 H2
94 g fenol..........23 g sodiu
a g fenol..............x g sodiu
94x=23a=>a=94x/23
46 g etanol.............23 g sodiu
b g etanol................y g sodiu
23b=46y=>b=2y
a+b=23.4
x+y=6.9
94x/23+2y=23.4
x+y=6.9=>x=6.9-y
94*(6.9-y)/23+2y=23.4
28.2-94y/23+2y=23.4 (*23)
648.6-94y+46y=538.2=>48y=110.4=>y=2,3 g Na
x = 6.9 - 2.3 = 4.6 g sodiu
a = 94*4.6/23 = 18.8 g fenol
Cu NaOH reactioneaza doar fenolul
94 g fenol...........40 g NaOH
18.8 g fenol...............x g NaOH
x = 8 g NaOH = 0.25 moli
cM = n/V => V = n/cM = 0.25/0.25 = 1 l

Answer 2

MCH3-CH2-OH=46g/mol
MC6H5-OH=94g/mol

CH3-CH2-OH + Na ---> CH3-CH2-ONa (etoxid de sodiu) + 1/2H2

C6H6-OH + Na ---> C6H5-ONa (fenoxid de sodiu) + 1/2H2

46x+94y=23.4        ==>46x+94y=23.4
23x+23y=6.9  ||(-2)==>-46x-46y=-13.8
                                      /      48y=9.6==>y=0.2 moli

46x+94*0.2=23.4==>46x+18.8=23.4==>46x=4.6==>x=0.1 moli

m alcool=46*0.1=4.6g

m fenol=0.2*94=18.8g

Cu NaOH reactioneaza doar fenolul :

94g...............40g
C6H5-OH + NaOH ---> C6H5-ONa + H2O
18.8g............x=8g

n=m/M=8/40=0.2 moli NaOH

Cm=n/Vs==>Vs=n/Cm=0.2/0.25=0.8L=800mL NaOH

MNaOH=40g/mol