Question

O arena monunucleara A,care contine 8,69%H formeaza prin clorurare la lumina un compus clorurat B care contine 44,1%Cl.determinati:
a)Formulele moleculare ale compusilor A si B
b)Cantitatea de compus B care reulta din41,4g hidrocarbura A,randamentul reactiei fiind de 80%.

Answer 1

a)
arene mononucleare: CnH2n-6
Marena=12n+2n-6=14n-6 g/mol
14n-6.................100%
2n-6..................8,69%
200n-600=121,66n-52,14
78,34n=547,86 => n=7 => C7H8 => C6H5-CH3 - toluenul
la clorurarea fotochimica obtine amestec de trei compusi clorurati, clorurile de benzil,benziliden si benzin
notam formula compusului: C6H5-CH3-nCln
 (C6H5-CH3 +  nCl2 -> C6H5-CH3-nCln + nHCl)
Mcompus=77+15-n+35,5n=92+34,5n
34,5n+92.....................100%
35,5n.........................44,1%
34,5n+92=35,5*100n/44,1=80,5n => 46n=92 => n=2
deci compusul este C6H5-CHCl2 - clorura de benziliden
b)
92g...........................161g
C6H5-CH3 + 2Cl2 -> C6H5-CHCl2 + 2HCl
41,4g........................72,45g
80/100=mpractica/72,45 => mpractica=57,96g