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\(n_{hhk}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{Br_2}=\dfrac{128.10}{160.100}=0,08\left(mol\right)\)
\(C_2H_4+Br_{2\left(dd\right)}\rightarrow C_2H_4Br_2\)
0,08 0,08 0,08 ( mol )
\(\left\{{}\begin{matrix}V_{C_2H_4}=0,08.22,4=1,792\left(l\right)\\V_{CH_4}=5,6-1,792=3,808\left(l\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,08}{0,25}.100=32\%\\\%V_{CH_4}=100-32=68\%\end{matrix}\right.\)
\(m_{C_2H_4Br_2}=0,08.188=15,04\left(g\right)\)
Cho hỗn hợp qua dung dịch brom chỉ có etylen tác dụng.
\(n_{Br_2}=0,25\cdot1,5=0,375mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,375 0,375
\(V_{C_2H_4}=0,375\cdot22,4=8,4l\Rightarrow V_{CH_4}=11-8,4=2,6l\)
a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(1\right)\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{48}{160}=0,3\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,2.22,4}{5,6}.100\%=80\%\\\%V_{C_2H_2}=20\%\end{matrix}\right.\)
b, \(V_{ddBr_2}=\dfrac{0,3}{0,25}=1,2\left(M\right)\)
a, PT: \(C_3H_6+Br_2\rightarrow C_3H_6Br_2\)
Ta có: m bình tăng = mC3H6 = 6,3 (g)
\(\Rightarrow n_{C_3H_6}=\dfrac{6,3}{42}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_3H_6}=\dfrac{0,15.22,4}{6,72}.100\%=50\%\\\%V_{C_2H_6}=100-50=50\%\end{matrix}\right.\)
b, Theo PT: \(n_{Br_2}=n_{C_3H_6}=0,15\left(mol\right)\Rightarrow C_{M_{Br_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)
c, Theo PT: \(n_{C_3H_6Br_2}=n_{C_3H_6}=0,15\left(mol\right)\Rightarrow C_{M_{C_3H_6Br_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)
