Question

O cantitate de fenol se nitreaza pana la trinitrofenol.La nitrare s-au utilizat 630 g solutie de acid azotic de concentratie 80%.La sfarsitul reactiei,dupa indepartarea trinitrofenolului,solutia de acid azotic are o concentratie de 35%.Ce cantitate de fenol s-a nitrat?

Answer 1

Cp=md/ms*100==>md=Cp*ms/100=504g HNO3
sol HNO3i=504g;  ms=md+mH2O==>mH2O=ms-md=630-504=126g H2O
sol HNO3f= mH2Oi + mH2O din reactie= 126+3x*18=124+54x g ; mHNO3reactionat=mHNO3i - HNO3 din reactie= 504-3x*53= 504-189x 
Cp=35%
35=504-189x/504-189x+126+54x*100=>50400-18900x=22050-4725x==>14175x=28350=>x=2 moli
mfenol=n*M=2*94=188g fenol

C6H5-OH + 3HO-NO2/H2SO4--> C6H3N3O7 (TNT-Acid Picric)+ 3H2O

Answer 2

C6H5 - OH + 3HONO2 ⇒ (NO2)₃C6H2 - OH + 3H2O
solutia initiala : ms = 630g    md = 504g = mi HNO3
au reactionat : n fenol + 3n HNO3  (63·3n g HNO3) ⇒ n trinitrofenol + 3n H2O
solutia reziduala de HNO3 : ms = 630 - 189n + 3n·18 = 630 -135n
c = 35%         md = 504 - 189n
(504 - 189n)·100 / (630 - 135n) = 35
50400 - 18900n = 22050 - 4725n
14175n = 28350    n fenol = 2moli   m fenol = 188g