1. CxH(2x - 6 ) M A= 14x - 6
precipitat alb : AgCl n AgCl = 2,87/143,5 = 0,02 moli = nA = nB
B : CxH(6x - 7)Cl M B = M A + 34,5
nB = 2,83 /(14x - 6 +35,5) = 0,02 14x + 29,5 = 141,5 14x = 112 x = 8
A : C8H10 formeaza un singur compus monoclorurat ⇒ A = para xilen
B : C8H9Cl
C6H5- CH2-CH3 ⇒ C6H5 - CH=CH2 +H2
2. R-Br + Mg ⇒ R-MgBr + H2O ⇒ R-H + MgBrOH
V teoretic B = 252·100/75 = 336ml nB = nA = 0,336/22,4 = 0,015 moli
M A = 2,055/0,015 = 137g/moli M R + 80 = 137 M R = 57
R = -C4H9 Br - CH2-CH2 - CH2 -CH3