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![](https://rs.olm.vn/images/avt/0.png?1311)
\(Gọi : n_{C_2H_4} = a; n_{C_2H_2} = b\\ \Rightarrow a + b = \dfrac{5,6}{22,4} = 0,25(1)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ n_{Br_2} = a + 2b = \dfrac{56}{160} =0,35(2)\\ (1)(2)\Rightarrow a = 0,15 ; b = 0,1\\ \Rightarrow \%V_{C_2H_4} = \dfrac{0,15}{0,25} .100\% = 60\%\\ \%V_{C_2H_2} = 100\% -60\% = 40\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi số mol của C2H4 và C2H2 lần lượt là x và y mol
theo bài ra: x+y = 0,56/22,4 = 0,025 (mol)
Pt:
C2H4 + Br2 → C2H4Br2
x mol x mol x mol
C2H2 + 2 Br2 → C2H2Br4
y mol 2y mol y mol
Số mol n Br2 = x+2y = 5,6/160 = 0,035 9mol)
Giải hệ ta đc: x = 0,015 và y = 0,01
=> %V C2H4 = 0,015/0,025 = 60% ; %V C2H2 = 40%
![](https://rs.olm.vn/images/avt/0.png?1311)
C2H4+Br2->C2H4Br2
x----------x---------x
C2H2+2Br2->C2H2Br4
y--------2y------------y
=>\(\left\{{}\begin{matrix}x+y=\dfrac{0,896}{22,4}\\160x+320y=8\end{matrix}\right.\)
=>x=0,03 mol, y=0,01 mol
=>%VC2H4=\(\dfrac{0,03.22,4}{0,896}\).100=75%
=>%VC2H2=25%
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3
a) C2H4 + Br2 --> C2H4Br2
C2H2 + 2Br2 --> C2H2Br4
b) \(n_{Br_2}=\dfrac{5,6}{160}=0,035\left(mol\right)\)
Gọi số mol C2H4, C2H2 là a, b (mol)
=> \(a+b=\dfrac{0,56}{22,4}=0,025\) (1)
PTHH: C2H4 + Br2 --> C2H4Br2
a---->a
C2H2 + 2Br2 --> C2H2Br4
b---->2b
=> a + 2b = 0,035 (2)
(1)(2) => a = 0,015 (mol); b = 0,01 (mol)
=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,015}{0,025}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{0,01}{0,025}.100\%=40\%\end{matrix}\right.\)
Bài 4:
a)
CH4 + 2O2 --to--> CO2 + 2H2O
2H2 + O2 --to--> 2H2O
b)
Gọi số mol CH4, H2 là a, b (mol)
=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (1)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
a-------------------->a--->2a
2H2 + O2 --to--> 2H2O
b--------------->b
=> \(2a+b=\dfrac{16,2}{18}=0,9\) (2)
(1)(2) => a = 0,4 (mol); b = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,4}{0,5}.100\%=80\%\\\%V_{H_2}=\dfrac{0,1}{0,5}.100\%=20\%\end{matrix}\right.\)
c)
VCO2 = 0,4.22,4 = 8,96 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\begin{array} {l} n_{Br_2}=\dfrac{41,6}{160}=0,26(mol)\\ \text{Đặt }n_{C_2H_4}=x(mol);n_{C_2H_2}=y(mol)\\ \to x+y=\dfrac{4,48}{22,4}=0,2(1)\\ C_2H_4+Br_2\to C_2H_4Br_2\\ C_2H_2+2Br_2\to C_2H_2Br_4\\ \text{Theo PT: }x+2y=n_{Br_2}=0,26(2)\\ (1)(2)\to\begin{cases} x=0,14\\ y=0,06 \end{cases} \\ \to \begin{cases} \%V_{C_2H_4}=\dfrac{0,14}{0,2}.100\%=70\%\\ \%V_{C_2H_2}=100-70=30\% \end{cases} \end{array}\)
\(\begin{array} {l} n_{Br_2}=\dfrac{41,6}{160}=0,26(mol)\\ \text{Đặt }n_{C_2H_4}=x(mol);n_{C_2H_2}=y(mol)\\ \to x+y=\dfrac{4,48}{22,4}=0,2(1)\\ C_2H_4+Br_2\to C_2H_4Br_2\\ C_2H_2+2Br_2\to C_2H_2Br_4\\ \text{Theo PT: }x+2y=n_{Br_2}=0,26(2)\\ (1)(2)\to\begin{cases} x=0,14\\ y=0,06 \end{cases} \\ \to \begin{cases} \%V_{C_2H_4}=\dfrac{0,14}{0,2}.100\%=70\%\\ \%V_{C_2H_2}=100-70=30\% \end{cases} \end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\left(1\right)\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}+n_{C_2H_2}=\dfrac{6,4}{160}=0,04\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,01\left(mol\right)\\n_{C_2H_2}=0,015\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,01.22,4}{0,56}.100\%=40\%\\\%V_{C_2H_2}=60\%\end{matrix}\right.\)
a)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b)
Gọi số mol C2H2, C2H4 là a, b
=> \(a+b=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{Br_2}=\dfrac{22,4}{160}=0,14\left(mol\right)\)
PTHH:\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
a---->2a
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b--->b
=> 2a + b = 0,14
=> a = 0,04; b = 0,06
\(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,04}{0,1}.100\%=40\%\\\%V_{C_2H_4}=\dfrac{0,06}{0,1}.100\%=60\%\end{matrix}\right.\)