Question

Determinati masa solutiei de HNO3 de concentratie C=63% necesara reactiei cu 846kg fenol, stiind ca raportul molar intre mononitrofenol:dinitrofenol:trintrofenol:fenol nereactionat=4:3:2:3

Answer 1

C6H5-OH + HO-NO2 => nitrofenol + H2O

C6H5-OH + 2HO-NO2 => dinitrofenol + 2H2O

C6H5-OH + 3HNO3 => acid picric + 3H2O

C6H5-OH => C6H5-OH (nereactionat)


n=m/M=846/94 = 9 kmoli C6H5-OH

x+y+z+w = 9

x/y = 4/3 => y=3x/4
x/z = 4/2 => z=2x/4
x/w = 4/3 => w=3x/4

x+3x/4+2x/4+3x/4 = 9 => 12x/4 = 9 => x=3 kmoli

y=2,25 kmoli

z=1,5 kmoli

w = 2,25 kmoli

x+2y+3z = 3+4,5+4,5 = 12 kmoli HNO3

md HNO3 12*63 = 756kg

C=md/ms*100=>ms=md*100/C = 1200kg sol. HNO3