n=m/M=32,4/81=0,4 moli HBr
CnH2n + HBr => CnH2n+1Br
14n...............1 mol HBr
16,8g.............0,4 moli HBr
n = 3 => F.M => C3H6 - propena
H3C-CH=CH2 + HBr =>>>> H3C-CH-CH3 -->>> [F.S]
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Br
Denumire: 2-bromopropan
nHBr=n C3H7Br = 0,4 moli
m C3H7Br = 0,4*123 = 49,2 g