F.M alchena => CnH2n
CnH2n + HBr => CnH2n+1Br
n=m/M = 32,4/81 = 0,4 moli HBr
MHBr = 81g/mol
14ng alchena............1 mol HBr
16,8g ....................0,4 moli HBr
5,6n = 16,8 => n=3
F.M alchena => C3H6 -> propena
F.M compus rezultat => H3C-CH(Br)-CH3, 2-bromopropan
m compus rezultat = 0,4n*M = 0,4*123 = 49,2g