Question

Prin nitrarea a 9.2 g de toluen cu amestec sulfonitric se formeaza compusul A ce contine 18.5% N. Masa amestecului sulfonitric ce contine 31.5%HNO3 necesar nitrarii este?

Answer 1

Ecuatia reactiei:
C7H8  +3HNO3-->C7H5(NO2)3   + 3H20
M = (7X12gC+8gH)/mol)=92g/mol toluen
M= (7x12gC +5gH+3X46gNO2)/mol= 227g/mol  TRINITROTOLUEN
Am verificat procentul de N, 
227g trinitrotoluen......3x14gN
100g.........................18,5

-revenind la ecuatie,
din 92g toluen...........227g produs................................3x 63gHNO3
     9,2g......................m= 22,7g trinitrotolurn.................m=18,9g HNO3
-calculez masa de amestec sulfo-nitric
100g amestec.......contine 31,5g  HNO3
m=......................................18,9g
CALCULEAZA !!!!!