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\(n_{hh}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,05 0,05 ( mol )
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,5}.100=10\%\\\%V_{CH_4}=100\%-10\%=90\%\end{matrix}\right.\)
--> Chọn B
Ta có: m bình Brom tăng = mC2H4 = 5,6 (g)
\(\Rightarrow n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\Rightarrow\%V_{C_2H_4}=\dfrac{0,2.22,4}{5,6}.100\%=80\%\)
\(\Rightarrow\%V_{CH_4}=100-80=20\%\)
\(\left\{{}\begin{matrix}C_2H_4:x\left(mol\right)\\C_2H_2:y\left(mol\right)\end{matrix}\right.\)⇒ x + y = \(\dfrac{6,72}{22,4}=0,3\left(1\right)\)
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\)
Theo PTHH :
x + 2y = \(\dfrac{64}{160} = 0,4(2)\)
Từ (1)(2) suy ra: x = 0,2 ; y = 0,1
Vậy :
\(\%V_{C_2H_4} = \dfrac{0,2}{0,3}.100\% = 66,67\%\\ \%V_{C_2H_2} = 100\% - 66,67\% = 33,33\%\)
\(n_{CO_2}=0.3\left(mol\right)\)
\(Đặt:n_{C_2H_2}=a\left(mol\right),n_{C_2H_4}=b\left(mol\right)\)
\(n_{Br_2}=\dfrac{64}{160}=0.4\left(mol\right)\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(\left\{{}\begin{matrix}a+b=0.3\\2a+b=0.4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.2\end{matrix}\right.\)
\(\%V_{C_2H_2}=\dfrac{0.1}{0.3}\cdot100\%=33.33\%\)
\(\%V_{C_2H_4}=66.67\%\)
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = \dfrac{4}{160}=0,025(mol)\\ \%V_{C_2H_4} = \dfrac{0,025.22,4}{5,6}.100\% = 10\%\\ \%V_{CH_4}= 100\%-10\%=90\%\)
a) Các phương trình phản ứng:
C2H4 + Br2 → C2H4Br2
C2H2 + 2Br2 → C2H2Br4
C2H2 + Ag2O → C2Ag2 + H2O
Hay
C2H2 + 2AgNO3 + 2NH3 → C2Ag2 + 2NH4NO3
b) Gọi a, b, c lần lượt là số mol của CH4, C2H4, C2H2 trong 4,3gam hỗn hợp T.
– Số mol Br2 = 0,15 (mol); số mol kết tủa = số mol C2H2 = 0,075 (mol); số mol T = 0,3 (mol). Do đó nT = 4nC2H2
– Ta có hệ phương trình:
– Suy ra % thể tích mỗi khí trong T:
%VCH4 = 50%; %VC2H2 = %VC2H4 = 25%
\(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right);n_{hh}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,025<-0,125
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,025}{0,125}.100\%=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)
Ta có: \(n_{Br_2}=\dfrac{1,6}{160}=0,01\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,01\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,01.22,4}{0,5}.100\%=44,8\%\)
\(\Rightarrow\%V_{CH_4}=100-44,8=55,2\%\)
\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{56}{160}=0,35mol\)
Gọi \(n_{C_2H_4}\) là x \(\Rightarrow V_{C_2H_4}=22,4x\)
\(n_{C_2H_2}\) là y \(\Rightarrow V_{C_2H_2}=22,4y\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
x x ( mol )
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
y 2y ( mol )
Ta có:
\(\left\{{}\begin{matrix}22,4x+22,4y=5,6\\x+2y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(\Rightarrow V_{C_2H_4}=22,4.0,15=3,36l\)
\(\Rightarrow V_{C_2H_2}=22,4.0,1=2,24l\)
\(\%V_{C_2H_4}=\dfrac{3,36}{5,6}.100=60\%\)
\(\%V_{C_2H_2}=\dfrac{2,24}{5,6}.100=40\%\)
nhh khí = 5,6/22,4 = 0,25 (mol)
Gọi nC2H4 = a (mol); nC2H2 = b (mol)
a + b = 0,25 (1)
nBr2 = 56/160 = 0,35 (mol)
PTHH:
C2H4 + Br2 -> C2H4Br2
Mol: a ---> a
C2H2 + 2Br2 -> C2H2Br4
Mol: b ---> 2b
a + 2b = 0,35 (2)
(1)(2) => a = 0,15 (mol); b = 0,1 (mol)
%VC2H2 = 0,15/0,25 = 60%
%VC2H4 = 100% - 60% = 40%
1) \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{1,68}{6,72}\cdot100\%=25\%\\\%V_{C_2H_2}=75\%\end{matrix}\right.\)
2) Ta có: \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,2}{\dfrac{5,6}{22,4}}\cdot100\%=80\%\) \(\Rightarrow\%V_{CH_{_4}}=20\%\)
huhuu bạn nào giúp tớ với ....