nC2H4Br2 = \(\dfrac{48}{188}\approx0,255\left(mol\right)\)
a) C2H4 + Br2 -> C2H4Br2
0,255 <-- 0,255 <-- 0,255
b) mBr2 = 0,255. 160 = 40.8 (g)
c) VC2H4 = 0,255 . 22,4 = 5,712 (l)
VCH4 = 11,2 - 5,712 = 5,488 (l)
%VCH4 = \(\dfrac{5,488.100}{11,2}=49\%\)
%VC2H4 = \(\dfrac{5,712.100}{11,2}=51\%\)