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156 g benzen se supun reactiei de bromurare, folosind o solutie 20% de brom. Se cere:
a) ecuatia reactiei chimice de monobromurare
b) masa de solutie de brom folosita
c) masa de produs rezultat la randament de 75%
PS: Va rog sa ma ajutati macar la a si la b.

Răspuns :

a) C6H6 + Br-Br --> C6H5-Br + HBr

b) MC6H6=78g/mol
MBr2=160g/mol

n=m/M=156/78=2 moli C6H6

1mol C6H6......1mol Br2
2mol...................2mol

n=md/M==>md=n*M=2*160=320g Br2

Cp=md/ms*100==>ms=md*100/20=320*100/20=1600g Br2

c) 1mol..........1mol.........1mol
     2mol..........2mol.........2mol

MC6H5-Br=12*6+5+80=157g/mol

n=md/M==>md=n*M=2*157=314g (mt)

n(eta)=mp/mt*100==>mp=n*mt/100=75*314/100=235.5g bromobenzen

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