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a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{C_2H_4}=n_{Br_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{3,36}.100\%\approx66,67\%\\\%V_{CH_4}\approx33,33\%\end{matrix}\right.\)
Ta có: m dd Br2 tăng = mC2H4 = 2,8 (g)
\(\Rightarrow n_{C_2H_4}=\dfrac{2,8}{28}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{3,36}.100\%\approx66,67\%\\\%V_{CH_4}\approx33,33\%\end{matrix}\right.\)
Có: \(n_{CH_4}=\dfrac{3,36}{22,4}-0,1=0,05\left(mol\right)\)
⇒ m hh = mCH4 + mC2H4 = 0,05.16 + 0,1.28 = 3,6 (g)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,05.16}{3,6}.100\%\approx22,22\%\\\%m_{C_2H_4}\approx77,78\%\end{matrix}\right.\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Gọi: \(\left\{{}\begin{matrix}n_{C_2H_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow x+y=\dfrac{2,24}{22,4}=0,1\left(mol\right)\left(1\right)\)
\(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=x+2y=\dfrac{24}{160}=0,15\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x=y=0,05\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,05.22,4}{2,24}.100\%=50\%\)
\(n_{Br_2}=\dfrac{8}{160}=0.05\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(0.05......0.05\)
\(V_{C_2H_4}=0.05\cdot22.4=1.12\left(l\right)\)
\(V_{CH_4}=20-1.12=18.88\left(l\right)\left(mol\right)\)
\(\%V_{C_2H_4}=\dfrac{1.12}{20}\cdot100\%=5.6\%\)
\(\%V_{CH_4}=100-5.6=94.4\%\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,1<--0,1
=> \(\%V_{C_2H_4}=\dfrac{0,1.22,4}{22,4}=10\%\)
=> %VCH4 = 100% - 10% = 90%
\(n_{hh}=\dfrac{4,48}{22,4}=0,2mol\)
\(n_{Br_2}=\dfrac{48}{160}=0,3mol\)
Gọi \(\left\{{}\begin{matrix}n_{C_2H_2}=x\\n_{C_2H_4}=y\end{matrix}\right.\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
x 2x ( mol )
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}x+y=0,2\\2x+y=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,1}{0,2}.100=50\%\\\%V_{C_2H_4}=100\%-50\%=50\%\end{matrix}\right.\)
\(n_{Br_2} = \dfrac{4,8}{160} = 0,03(mol)\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ n_{C_2H_2}= \dfrac{1}{2}n_{Br_2} = 0,015(mol)\\ \%V_{C_2H_2} = \dfrac{0,015.22,4}{1,12}.100\% = 30\%\\ \%V_{CH_4} = 100\% -30\%=70\%\)