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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
\(V_{C_2H_4}=0,375\cdot22,4=8,4l\Rightarrow V_{CH_4}=11-8,4=2,6l\)
\(n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\\ n_{hh2khi}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ ChỉcóC_2H_4tácdụngvớiBr_2\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4}=n_{Br_2}=0,4\left(mol\right)\\ \Rightarrow n_{CH_4}=0,5-0,4=0,1\left(mol\right)\\ \%Vcũnglà\%n\\ \Rightarrow\%V_{CH_4}=\dfrac{0,1}{0,5}.100=20\%;\%V_{C_2H_4}=100-20=80\%\\ CM_{Br_2}=\dfrac{0,4}{0,25}=1,6M\)
\(n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\)
\(n_{hh}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,4<----0,4
\(\rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,4}{0,5}.100\%=80\%\\\%V_{CH_4}=100\%-80\%=20\%\end{matrix}\right.\)
\(C_{M\left(Br_2\right)}=\dfrac{0,4}{0,25}=1,6M\)
\(n_{hhk}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) ; \(n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\)
\(C_2H_4+Br_{2\left(dd\right)}\rightarrow C_2H_4Br_2\)
0,4 0,4 ( mol )
\(\left\{{}\begin{matrix}V_{C_2H_4}=0,4.22,4=8,96\left(l\right)\\V_{CH_4}=11,2-8,96=2,24\left(l\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,4}{0,5}.100=80\%\\\%V_{CH_4}=100-80=20\%\end{matrix}\right.\)
\(C_{M_{Br_2}}=\dfrac{0,4}{0,25}=1,6\left(M\right)\)
Cho hỗn hợp qua dung dịch brom dư
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Khí thoát ra là \(CH_4\)
\(CH_4+2O_2\rightarrow^{t^o}CO_2+2H_2O\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
Ta có:
\(n_{CaCO_3}=\frac{40}{100}=0,4mol=n_{CO_2}=n_{CH_4}\)
\(\rightarrow V_{CH_4}=0,4.22,4=8,96l\)
\(\rightarrow\%V_{CH_4}=\frac{8,96}{13,56}=66\%\rightarrow\%V_{C_2H_4}=34\%\)
nhh = 5.6/22.4 = 0.25 (mol)
nBr2 = 0.5*0.2 = 0.1 (mol)
C2H4 + Br2 => C2H4Br2
0.1_____0.1
nCH4 = 0.25 - 0.1 = 0.15 (mol)
VC2H4 = 0.1*22.4 = 2.24 (l)
VCH4 = 0.15*22.4 = 3.36 (l)
mCH4 = 0.15*16 = 2.4 (g)
mC2H4 = 0.1*28 = 2.8 (g)
%mCH4 = 2.4/(2.4 + 2.8) * 100% = 46.15%
%mC2H4 = 100 - 46.15 = 53.85%
Chúc bạn học tốt !!!
\(n_{hhk}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{Br_2}=\dfrac{128.10}{160.100}=0,08\left(mol\right)\)
\(C_2H_4+Br_{2\left(dd\right)}\rightarrow C_2H_4Br_2\)
0,08 0,08 0,08 ( mol )
\(\left\{{}\begin{matrix}V_{C_2H_4}=0,08.22,4=1,792\left(l\right)\\V_{CH_4}=5,6-1,792=3,808\left(l\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,08}{0,25}.100=32\%\\\%V_{CH_4}=100-32=68\%\end{matrix}\right.\)
\(m_{C_2H_4Br_2}=0,08.188=15,04\left(g\right)\)