Ta có:
\(n_{hh}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{Br2}=100.10^{-3}.1,5=0,15\left(mol\right)\)
=> Etilen tác dụng với Br2 theo tỉ lệ 1:1
\(\Rightarrow n_{etilen}=n_{Br2}=0,15\left(mol\right)\)
\(n_{CH4}=0,25-0,15=0,1\left(mol\right)\)
\(\%V_{CH4}=\frac{0,1}{0,25}.100\%=40\%\)
\(\%V_{C2H4}=100\%-40\%=60\%\)
\(\Rightarrow m_{hh\left(khi\right)}=0,1.16+0,15.28=5,8\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH4}=\frac{0,1.16}{5,8}.100\%=28\%\\\%m_{C2H4}=100\%-28\%=72\%\end{matrix}\right.\)
\(\Rightarrow m_{C2H4Br2}=0,15.188=28,2\left(g\right)\)
\(C^{4-}\rightarrow C^{4+}+8e\)
0,1__________ 0,8
\(2C^{2-}\rightarrow2C^{4+}+12e\)
0,15___0,15_____1,8
\(O_2\rightarrow2O_2^-+4e\)
x____________4x
\(4x=1,8+0,8\Rightarrow x=6,5\left(mol\right)\)
Vì O2 chiếm 20% thế tích ko khí nên:
\(\Rightarrow V_{kk}=\frac{22,4.6,5.100}{20}=728\left(l\right)\)
