Question

Se barboteaza intr o sol. de H2O de Br de conc. 1 %, 2.24 ml propina.Pt. decolorarea totala a apei de Br se mai barboteaza inca 13.44 cm^3 amestec echimolecular de propan si propena.Se cere:A.scrie ec. react. chimice care au avut loc
B.calculeaza masa sol. de apa de brom

Answer 1

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