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\(n_{hh}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,05 0,05 ( mol )
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,5}.100=10\%\\\%V_{CH_4}=100\%-10\%=90\%\end{matrix}\right.\)
--> Chọn B
Ta có: m bình Brom tăng = mC2H4 = 5,6 (g)
\(\Rightarrow n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\Rightarrow\%V_{C_2H_4}=\dfrac{0,2.22,4}{5,6}.100\%=80\%\)
\(\Rightarrow\%V_{CH_4}=100-80=20\%\)
1) \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{1,68}{6,72}\cdot100\%=25\%\\\%V_{C_2H_2}=75\%\end{matrix}\right.\)
2) Ta có: \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,2}{\dfrac{5,6}{22,4}}\cdot100\%=80\%\) \(\Rightarrow\%V_{CH_{_4}}=20\%\)
\(Đặt:n_{CH_4}=a\left(mol\right);n_{C_2H_4}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,25\\2a+3b=0,625\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,125\\b=0,125\end{matrix}\right.\\ a,m_{hh}=m_{CH_4}+m_{C_2H_4}=16.0,125+28.0,125=5,5\left(g\right)\\ b,V_{CO_2\left(đktc\right)}=22,4.\left(a+2b\right)=8,4\left(l\right)\)
Ta có: \(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)
\(\Rightarrow V_{CH_4}=7,72-4,48=3,24\left(l\right)\)