Theo đề cho : nBr2 = \(\dfrac{16}{160}\) = 0.1(mol)
PTHH : C2H4 + Br2 \(\rightarrow\) C2H4Br2
0.1 \(\leftarrow\) 0.1 (mol)
\(\Rightarrow\) VC2H4 = 0.1 \(\times\) 22.4 = 2.24(lít)
Vậy : %VC2H4 = \(\dfrac{2.24\times100\%}{11.2}\)= 20%
%VCH4 = 100% - 20% = 80%