1. n Br2 = 0,0625(mol)
Metan không pứ với ddBr2
PTHH: C2H4 + Br2 -> C2H4Br2
(mol)___0,0625__0,0625________
%VC2H4 = (22,4.0,0625)/5,6.100%=25%
=> %VCH4 = 100-25=75%
2.
n C6H5Br = 0,25(mol)
PTHH: C6H6 + Br2 -to,xt-> C6H5Br+HBr
(mol)____0,25________________0,25_________
m C6H6 = 0,25.78=19,5(g)
Do H = 80%
=> m C6H6 = 19,5:80%=24,375(g)