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a)
Khí thoát ra: CH4
\(\%V_{CH_4} = \dfrac{6,72}{16,8}.100\% = 40\%\\ \%V_{C_2H_4} = 100\% - 40\% = 60\%\)
b)
\(n_{C_2H_4} = \dfrac{16,8-6,72}{22,4} = 0,45(mol)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{Br_2} = n_{C_2H_4} = 0,45(mol)\\ \Rightarrow C_{M_{Br_2}} = \dfrac{0,45}{2} = 0,225M\\ c) n_{C_2H_4Br_2} = n_{C_2H_4} = 0,45(mol)\\ \Rightarrow n_{C_2H_4Br_2} = 0,45.188 = 84,6(gam)\)
Bài 4:
a) n(hỗn hợp khí)= 16,8/22,4=0,75(mol)
- Khí thoát ra là khí CH4.
=> nCH4=6,72/22,4=0,3(mol)
nC2H4=0,75-0,3=0,45(mol)
- Số mol tỉ lệ thuận với thể tích.
%V(CH4)=%nCH4= (0,3/0,75).100=40%
=> %V(C2H4)=100% - 40%=60%
b) PTHH: C2H4 + Br2 -> C2H4Br2
nC2H4Br2= nBr2=nC2H4=0,45(mol)
=>VddBr2= 0,45/2=0,225(l)
c) mC2H4Br2=0,45. 188= 84,6(g)
a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, \(n_{C_2H_4Br_2}=\dfrac{1,88}{188}=0,01\left(mol\right)\)
Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,01\left(mol\right)\Rightarrow V_{C_2H_4}=0,01.22,4=0,224\left(l\right)\)
\(\Rightarrow V_{CH_4}=20-0,224=19,776\left(l\right)\)
c, \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,224}{20}.100\%=1,12\%\\\%V_{CH_4}=98,88\%\end{matrix}\right.\)
d, \(n_{Br_2}=n_{C_2H_4Br_2}=0,01\left(mol\right)\Rightarrow C_{M_{Br_2}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\)
a) \(n_{C_2H_4Br_2}=\dfrac{37,6}{188}=0,2\left(mol\right)\)
PTHH: C2H4 + Br2 ---> C2H4Br2
0,2<---0,2<------0,2
b) \(\left\{{}\begin{matrix}V_{C_2H_4}=0,2.24,79=4,958\left(l\right)\\V_{CH_4}=20-4,958=15,042\left(l\right)\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{4,958}{20}.100\%=24,79\%\\\%V_{CH_4}=100\%-24,79\%=75,21\%\end{matrix}\right.\)
d) \(V_{\text{dd}Br_2}=\dfrac{0,2}{0,2}=1\left(l\right)\)
a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b, \(n_{C_2H_4Br_2}=\dfrac{37,6}{188}=0,2\left(mol\right)\)
\(n_{C_2H_4}=n_{C_2H_4Br_2}=0,2\left(mol\right)\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)
\(\Rightarrow V_{CH_4}=20-4,48=15,52\left(l\right)\)
c, \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{4,48}{20}.100\%=22,4\%\\\%V_{CH_4}=77,6\%\end{matrix}\right.\)
d, \(n_{Br_2}=n_{C_2H_4Br_2}=0,2\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,2}{0,2}=1\left(l\right)\)
a) \(V_{CH_4}=0,6\left(l\right)\)
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,6}{1,2}.100\%=50\%\\\%V_{C_2H_4}=100\%-50\%=50\%\end{matrix}\right.\)
b) \(n_{C_2H_4}=\dfrac{1,2-0,6}{24}=0,025\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,025-->0,025
=> \(m_{Br_2}=0,025.160=4\left(g\right)\)
c)
\(n_{CH_4}=\dfrac{0,6}{24}=0,025\left(mol\right)\)
=> nH = 0,025.4 = 0,1 (mol)
\(n_{Cl_2}=\dfrac{0,72}{24}=0,03\left(mol\right)\)
=> nCl(thế H) = 0,03 (mol)
Do nH > nCl(thế H)
=> H không bị thế hoàn toàn bởi Cl
=> nHCl = 0,03 (mol)
=> mHCl = 0,03.36,5 = 1,095 (g)
\(a)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = \dfrac{80.10\%}{160} = 0,05(mol)\\ \Rightarrow \%m_{C_2H_4} = \dfrac{0,05.28}{2}.100\% = 70\%\\ \%m_{CH_4} = 100\% - 70\% = 30\%\)
\(b)\) Sản phẩm : Đibrom etan
\(n_{C_2H_4Br_2} = n_{Br_2} = 0,05(mol)\\ \Rightarrow m_{C_2H_4Br_2} = 0,05.188 = 9,4\ gam\)
\(n_{hh}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(m_{Br_2}=200\cdot\dfrac{20}{100}=40\left(g\right)\)
\(n_{Br_2}=\dfrac{40}{160}=0.25\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(0.25........0.25..........0.25\)
\(\)\(n_{C_2H_4}=n_{hh}=0.25\left(mol\right)\)
=> Sai đề
\(n_{hh}=6,72:22,4=0,3mol\\ C_2H_2+2Br_2->C_2H_2Br_4\\ C_2H_4+Br_2->C_2H_2Br_2\\ n_{Br_2}=0,4mol\\ n_{C_2H_2}=a;n_{C_2H_4}=b\\ a+b=0,3\\ 2a+b=0,4\\ a=0,2;b=0,1\\ \%V_{C_2H_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\ \%V_{C_2H_4}=33,33\%\)
1/2 hỗn hợp có 0,1 mol C2H2 và 0,05mol C2H4
\(BT.C:n_{CO_2}=2n_{C_2H_2}+2n_{C_2H_4}=0,3mol\\ n_{CaCO_3}=n_{CO_2}=0,3\\ m_{KT}=0,3.100=30g\)
\(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05mol\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,05 0,05 0,05 ( mol )
\(m_{Br_2}=0,05.160=8g\)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,25}.100=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)
1. \(n_{Br_2}=0,4.0,5=0,2\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)
2. \(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,05\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,05.22,4}{1,4}.100\%=80\%\)
\(\Rightarrow\%V_{CH_4}=100-80=20\%\)
nC2H4 = 0.56/22.4 = 0.025 (mol)
C2H4 + Br2 => C2H4Br2
0.025__0.025____0.025
mC2H4Br2 = 0.025*188 = 4.7 (g)
mBr2 = 0.025*160 = 4 (g)
C% Br2 = 4/150 * 100% = 2.67%
a. \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b.\(n_{C_2H_4}=\dfrac{n}{22,4}=\dfrac{0.56}{22.4}=0.025\)mol
\(n_{C_2H_4Br_2}=\dfrac{0,025.1}{1}=0,025mol\)
\(m_{C_2H_4Br_2}=n.M=260.0,025=6.5g\)