a)
C6H6 + Cl2 --FeCl3--> C6H4Cl2 + 2HCl
b)
100% ......................... 400g benzen impur
98% ........................... m.pu = 392g benzen pur
392g V L md g
C6H6 + Cl2 --FeCl3--> C6H4Cl2 + 2HCl
78 22,4 147
=> V = 392x22,4/78 = 112,6 L Cl2 (c.n.)
c)
=> md = 392x147/78 = 738,77g derivat obtinut la 100% teoretic... dar ptractic la 98% cat s-a obtinut
100% .,...................... 738,77g derivat teoretic
98% ........................... practic = 724 g derivat
d)
Fm. = C6H4Cl2
raport masic: avem 6x12g C + 4x1g H + 2x35,5g Cl
C: H: Cl: = 72: 4: 71
e)
M.C6H4Cl2 = 147 g/mol
147g derivat .......................... 2x35,5g Cl
100% ....................................... a% = 48,30% Cl
f)
toti atomii de carbon sunt tertiari
