Question

Prin nitrarea benzenui se obtine 800g amestec cu 60% nitrobenzen, 20% dinitrobenzen, 10% trinitrobenzen si 10% benzen nereactionat. Calculati masa de benzen necesara procesului.

Answer 1

C6H6+HONO2--->C6H5NO2 +H2O
        +2HONO2--->C6H4(NO2)2+2H2O
        +3HONO2---> C6H3(NO2)3 +3H20
C6H6-------------------->C6H6
M,C6H6= 78g/mol
M,nitrobenzen= 123g/mol
M,dinitrobenzen= 168g/mol
M,trinitrobenzen=213g/mol
100g amestec....60g nitro-........20 g dinitro-...10gtrinitro-........10g benzen
800g..................48g................16g...................8g.................8g

din ecuatii, se deduce
 78gbenzen........123g nitro
m=.......................48g;   m= 
 
78g benzen........168 dinitro-
m'=.......................16;   m'=

78g benzen.........213g trinitr0-
m"=.......................8g     m"=

total benzen---->   m+m'+m"+8