\(n_{\downarrow}=n_{C_2Ag_2}=\dfrac{2,4}{240}=0,01mol\)
\(\Rightarrow n_{C_2H_2}=0,01mol\)
Nếu cho hỗn hợp này qua \(ddBr_2\) \(0,025mol\):
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,01 0,02
\(\Rightarrow n_{Br_2\left(trongC_2H_4\right)}=0,025-0,02=0,005mol\)
\(\Rightarrow n_{C_2H_4}=0,005mol\)
\(\%V_{C_2H_4}=\dfrac{0,005}{0,025}\cdot100\%=20\%\)
\(\%V_{C_2H_2}=100\%-20\%=80\%\)