2.24mL=0.00224L
n=V/Vm=0.00224/22.4=0.0001 moli C3H4
Cu propina (C3H4) reactioneaza 2 moli H-Br
1mol...............2mol
CH≡C-CH3 + H-Br --> CH3-C(Br)2-CH3
0.0001mol.....0.0002mol
n=V/Vm=13.44/22.4=0.6mmol=0.0006moli (amestec => Propena=0.0003mol + Propan=0.0003mol)
CH3-CH2-CH3 + H-Br ≠ nu reactioneaza
1mol.....................1mol
CH2=CH-CH3 + H-Br --> CH3-CH(Br)-CH3
0.0003mol.........0.0003mol
In total reactioneaza 0.0005moli HBr ==> n=m/M==>m=n*M=0.0005*160=0.08g HBr
Cp=md/ms*100==>ms=md*100/Cp=0.08*100/1=8g sol. HBr
MBr2=2ABr=2*80=160g/mol