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Se barboteaza 44,8 ml propena intr-un vas cu apa de brom de concentratie 2 % . Aflati masa de apa de brom care poate fi decolorata.

Răspuns :

Moonlight01
CH₂=CH-CH₃ + HBr = CH₃-CH(Br)-CH₃

MC3H6=42mg/mmol
MHBr=81gmg/mmol

22,4mL C3H6......................... 81mg HBr
44,8mL....................................x=162mg HBr

Cp=md/ms*100=>ms=md*100/Cp=162*100/2=8100mg=8,1g HBr

Sudoku95
C3H6 + Br2 + H2O => CH3-CH(OH)-CH2Br + HBr
22,4 l propena...........160 g Br2
44.8 ml propena............x g Br2
x = 320 mg = 0.32 g Br2
c = md/ms * 100 => ms = md/c * 100 = 0.32/2 * 100 = 16 g solutie Br2

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