Question

scrie ecuatia reactiei chimice de obtinere a trinitroluenului din toluen. Care estw masa de acid azotic de concentratie 70% necesata obtinerii a 2 kg trinitrotoluen?

Answer 1

Notez "*" inmultire si "/" impartire 
C6H5-CH3 + 3HO-NO2 ->C7H5N3O6 + 3H2O
3*63 Kg HNO3.............227 Kg C7H5N3O6
x Kg HNO3..................2 Kg C7H5N3O6
x=378/227=1.665 Kg HNO3
n=m/M=2/227=8.81 moli C7H5N3O6
Cp=md/ms*100 =>md=Cp*ms/100=70*2/100=1.4 Kg HNO3 ?
3 moli HNO3...........1 mol C7H5N3O6
x moli HNO3............8.81 moli C7H5N3O6
x=3*8.81=26.43 moli HNO3
n=md/M =>md=n*M=26.43*63=1665.09 g
Cp=md/ms*100 =>ms=md*100/Cp=1665.09*100/70=2378.8 g HNO3
Nu stiu care e buna rezolvarea. Daca ai rezultate la final verifica..