Question

Ce masa de 2,4,6-tribromoanilina se poate obtine din 37,2 g de anilina si 208 g de brom

Answer 1

   1mol...........1.5mol...........1mol
C6H5NH2+3/2Br2----->C6H2Br3-NH2

M,anilina=93g/mol.........................................n,mol= 37,2g/93g/mol=0,4mol
M,derivat bromurat=330g/mol
M,Br2= 160g/mol...........................................n,mol= 208g/160g/mol=1,3mol
in reactie , intra 0,4mol anilina si 0,4x3/2=0,6mol Br2, (0,7mol ramane in  exces)
rezulta n=0,4mol derivat bromurat
 m=nxM---> m= 0,4molx338g/mol=.............................calculeaza !!!!!