Question

Answer 1

C2H4+Br2-to>C2H4Br2

0,03------0,03

n Br2=0,03 mol

=>%VC2H4=\(\dfrac{0,03.22,4}{1}100\)=67,2%

=>%VCH4=32,8%

Answer 2

\(n_{Br_2}=0,2.0,15=0,03mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

0,03        0,03                       ( mol )

( CH₄ không tác dụng với dd Br₂ )

\(V_{C_2H_4}=0,03.22,4=0,672l\)

\(\rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,672}{1}.100=67,2\%\\\%V_{CH_4}=100\%-67,2\%=32,8\%\end{matrix}\right.\)