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a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
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: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{5,6}{160}=0,035\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,015\left(mol\right)\\n_{C_2H_2}=0,01\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{C_2H_4}=0,015.22,4=0,336\left(l\right)\\V_{C_2H_2}=0,01.22,4=0,224\left(l\right)\end{matrix}\right.\)
b, \(V_{ddBr_2}=\dfrac{0,035}{0,2}=0,175\left(l\right)\)
a, Ta có \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,05\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,05.22,4=1,12\left(l\right)\\V_{C_2H_4}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)
b, \(m_{CH_4}=0,05.16=0,8\left(g\right)\)
c, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)
\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)
Ta có Cùng điều kiện -> Quy số lít về số mol.n(hh ban đầu) = 20 mol; n(hh sau) = 16 lít
=> H2 phản ứng mất 4 lít => C2H2 có 2 lít và CH4 có 8 lít
\(V_{khí.giảm}=V_{C_2H_2}\)
\(\%C_2H_2=\dfrac{4,48}{6,72}.100=66,67\%\)
\(\%CH_4=100\%-66,67\%=33,33\%\)
\(\left\{{}\begin{matrix}CH_4:x\left(mol\right)\\C_2H_2:y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}16x+26y=11,6\\x+y=\dfrac{11,2}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,14mol\\y=0,36mol\end{matrix}\right.\)
a)\(\%V_{CH_4}=\dfrac{0,14}{0,5}\cdot100\%=28\%\)
\(\%V_{C_2H_2}=100\%-28\%=72\%\)
\(\%m_{CH_4}=\dfrac{0,14\cdot16}{11,6}\cdot100\%=19,31\%\)
\(\%m_{C_2H_2}=100\%-19,31\%=80,69\%\)
b)\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
Dẫn dung dịch qua bình đựng brom chỉ có \(C_2H_2\) tác dụng.
\(\Rightarrow m_{tăng}=m_{C_2H_2}=0,25\cdot26=6,5g\)
Bài 1:
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,05\left(mol\right)\)
\(\Rightarrow V_{C_2H_4}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow V_{CH_4}=3,36-1,12=2,24\left(l\right)\)
Bài 2:
Ta có: \(n_{C_2H_2}+n_{C_2H_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(1\right)\)
m dd tăng = mC2H2 + mC2H4
\(\Rightarrow6,8=26n_{C_2H_2}+28n_{C_2H_4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_2}=0,1\left(mol\right)\\n_{C_2H_4}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{C_2H_2}=0,1.22,4=2,24\left(l\right)\\V_{C_2H_4}=0,15.22,4=3,36\left(l\right)\end{matrix}\right.\)
Gọi số mol C2H4, C2H2 là a, b (mol)
=> a + b = \(\dfrac{13,44}{22,4}=0,6\) (1)
nBr2 = 0,8.1 = 0,8 (mol)
PTHH: C2H4 + Br2 --> C2H4Br2
a----->a
C2H2 + 2Br2 --> C2H2Br4
b----->2b
=> a + 2b = 0,8 (2)
(1)(2) => a = 0,4 (mol); b = 0,2 (mol)
=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,4}{0,6}.100\%=66,67\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,6}.100\%=33,33\%\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\) ( mol ) \(\Rightarrow m_{hh}=16x+28y=6\left(g\right)\) (1)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x x ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
y 2y ( mol )
\(n_{CO_2}=x+2y=0,4\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%m_{CH_4}=\dfrac{0,2.16}{6}.100=53,33\%\)
\(\%m_{C_2H_4}=100-53,33=46,67\%\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,1 0,1 ( mol )
\(m_{Br_2}=0,1.160=16\left(g\right)\)
a) Gọi số mol CH4, C2H4 là a, b (mol)
=> \(\left\{{}\begin{matrix}16a+28b=11,6\\a+b=\dfrac{11,2}{22,4}=0,5\end{matrix}\right.\)
=> a = 0,2 (mol); b = 0,3 (mol)
\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2}{0,5}.100\%=40\%\\\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,2.16}{11,6}.100\%=27,586\%\\\%m_{C_2H_4}=\dfrac{0,3.28}{11,6}.100\%=72,414\%\end{matrix}\right.\)
b)
\(n_{C_2H_4}=\dfrac{5,6.60\%}{22,4}=0,15\left(mol\right)\)
mtăng = mC2H4 = 0,15.28 = 4,2 (g)
PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có: \(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,025\left(mol\right)\)
\(\Rightarrow V_{C_2H_2}=0,025.22,4=0,56\left(l\right)\)
\(\Rightarrow V_{CH_4}=3,36-0,56=2,8\left(l\right)\)