6/.
metilbenzen : dimetilbenzen : benzen = 1:2:3
miu = 92 g/mol : 106 g/mol : 78 g/mol
n moli n moli
C6H6 + CH3Cl --> C6H5-CH3 + HCl
1 1
2n moli 2n moli
C6H6 + 2CH3Cl --> C6H4(CH3)2 + 2HCl
1 1
3n moli 3n moli
C6H6 ------------------> C6H6 netransformat
1 1
92n + 2n106 + 3n78 = 86,4 g
=> n = 0,16
=> 6n78 = 75,58 g =>A.
7/.
300g benzen impur ................ 100%
m g .............................................. 80% pur
= 240 g
240 g n g
C6H6 + ClCOCH3 --> C6H5COCH3 + HCl
78 36,5
=> n = 112,31 g HCl s-ar fi obtinut teoretic .... dar la noi s-au obtinut 106,5 g
=>
100% ....................... 112,31g HCl
eta% ....................... 106,5 g HCl
= 94,83% => B.
8/.
nitro-benzen + Cl2 --> m-cloronitrobenzen
C6H4NO2Cl, miu = 157,5 g/mol
=> %Cl = 35,5x100/157,5 = 22,54%
=> B.
