consideram ca din HNO3 se consuma x moli
C6H5-OH + 3HNO3 --------> TNF + 3H2O
1mol 3
moli 1 mol 3 moli
x/3 moli x moli x moli
md HNO3=315*80/100= 252 g
m sol finala de acid
=315 -63x + 18x= 315 - 45x
md fin=252-63x
35/100= (252-63x) / (315- 45x ) ==> x=3 moli
deci in reactie se
consuma un mol de fenol ( 94g)