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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_{Br_2}=\dfrac{6}{160}=0,0375\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,0375\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,0375.22,4}{6,72}.100\%=12,5\%\\\%V_{CH_4}=87,5\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a) C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = n_{C_2H_4Br_2} = \dfrac{18,8}{188} = 0,1(mol)\\ V_{C_2H_4} = 0,1.22,4 = 2,24(lít)\\ V_{CH_4} = 5,6 -2,24 = 3,36(lít)\\ b) m_{Br_2} = 0,1.180 = 18(gam) \)
![](https://rs.olm.vn/images/avt/0.png?1311)
5. Sửa đề: 6,72 l
nhh khí = 6,72/22,4 = 0,3 (mol)
nC2H4Br2 = 37,6/188 = 0,2 (mol)
PTHH: C2H4 + Br2 -> C2H4Br2
Mol: 0,2 <--- 0,2 <--- 0,2
%VC2H4 = 0,2/0,3 = 66,67%
%VCH4 = 100% - 66,67% = 33,33%
3. Dẫn qua Cl2 dư ngoài ánh sáng:
- Làm mất màu vàng của Cl2 -> CH4
CH4 + Cl2 -> (askt) CH3Cl + HCl
- Không hiện tượng -> CO2, C2H4 (*)
Dẫn các chất (*) qua dd Br2 dư:
- Làm Br2 mất màu -> C2H4
- Không hiện tượng -> CH4
4. M(hidrocacbon) = 2 . 14 = 28 (g/mol)
mC = 85,7% . 28 = 24 (g)
=> nC = 24/12 = 2 (mol)
mH = 28 - 24 = 4 (g)
=> nH = 4/1 = 4 (mol)
CTPT: C2H4
CTCT: CH2=CH2
![](https://rs.olm.vn/images/avt/0.png?1311)
C2H2+2Br2->C2H2Br4
x-----------2x
C2H4+Br2->C2H4Br2
y-----------2y
n Br2=0,8 mol
\(\left\{{}\begin{matrix}x+y=0,6\\2x+y=0,8\end{matrix}\right.\)
=>x=0,2 ,y=0,4 mol
=>%VC2H2=\(\dfrac{0,2.22,4}{13,44}100\)=33,33%
=>%C2H4=66,67%
C2H4+3O2-tO>2CO2+2H2O
C2H2+5\2O2-to>2CO2+H2O
=>Vkk=1,7.22,4.5=190,4l
![](https://rs.olm.vn/images/avt/0.png?1311)
a.\(m_{dd.Br_2\left(tăng\right)}=m_{C_2H_2}=2,6g\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)
\(\%V_{C_2H_2}=\dfrac{0,1}{0,25}.100=40\%\)
\(\%V_{CH_4}=100\%-40\%=60\%\)
b.\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,1 0,2 ( mol )
\(C_{M\left(dd.Br_2\right)}=\dfrac{0,2}{0,1}=2M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
nhh = 0,125(mol)
=> nC2H4Br2 = 4,7/188 = 0,025(mol)
C2H4 + Br2 => C2H4Br2
0,025_______0,025__________
=> nCH4 = 0,125-0,025 = 0,1(mol)
=> %VCH4 = 0,1.100/0,125 = 80%
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)
\(16n_{CH_4}+28n_{C_2H_4}=3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,1\left(mol\right)\\n_{C_2H_4}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%\approx53,33\%\\\%m_{C_2H_4}\approx46,67\%\end{matrix}\right.\)
- Ở cùng điều kiện nhiệt độ và áp suất, % số mol cũng là %V.
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)
b, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Có: m tăng = mC2H4 = 0,05.28 = 1,4 (g)
a) \(n_{hh}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=0,15\\16a+28b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%=66,67\%\\\%V_{C_2H_4}=100\%-66,67\%=33,33\%\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%=53,33\%\\\%m_{C_2H_4}=100\%-53,33\%=46,67\%\end{matrix}\right.\)
b) \(m=m_{C_2H_4}=0,05.28=1,4\left(g\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Ta có: \(n_{C_2H_4Br_2}=\dfrac{28,2}{188}=0,15\left(mol\right)\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=n_{C_2H_4Br_2}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,15.22,4}{5,6}.100\%=60\%\\\%V_{CH_4}=40\%\end{matrix}\right.\)
\(m_{Br_2}=0,15.160=24\left(g\right)\)