Cho 8,96 lít hỗn hợp khí C2H4 và C2H2 (đktc) phản ứng vùa đủ với 250ml dung dịch brom 2M.
a) Tính thành phần phần trăm thể tích mỗi khi trong hỗn hợp đầu.
b) Tính khối lượng các chất sau phản ứng
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,04}{0,1}.100\%=40\%\\\%V_{C_2H_4}=\dfrac{0,06}{0,1}.100\%=60\%\end{matrix}\right.\)
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) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,01\left(mol\right)\\n_{C_2H_2}=0,015\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,01.22,4}{0,56}.100\%=40\%\\\%V_{C_2H_2}=60\%\end{matrix}\right.\)
\(Gọi : n_{C_2H_4} = a; n_{C_2H_2} = b\\ \Rightarrow a + b = \dfrac{5,6}{22,4} = 0,25(1)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ n_{Br_2} = a + 2b = \dfrac{56}{160} =0,35(2)\\ (1)(2)\Rightarrow a = 0,15 ; b = 0,1\\ \Rightarrow \%V_{C_2H_4} = \dfrac{0,15}{0,25} .100\% = 60\%\\ \%V_{C_2H_2} = 100\% -60\% = 40\%\)
a)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,05<--0,05
=> \(V_{C_2H_4}=0,05.22,4=1,12\left(l\right)\)
=> \(V_{CH_4}=4,48-1,12=3,36\left(l\right)\)
b) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{1,12}{4,48}.100\%=25\%\\\%V_{CH_4}=\dfrac{3,36}{4,48}.100\%=75\%\end{matrix}\right.\)
\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{56}{160}=0,35mol\)
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%
\(\begin{array} {l} n_{Br_2}=\dfrac{41,6}{160}=0,26(mol)\\ \text{Đặt }n_{C_2H_4}=x(mol);n_{C_2H_2}=y(mol)\\ \to x+y=\dfrac{4,48}{22,4}=0,2(1)\\ C_2H_4+Br_2\to C_2H_4Br_2\\ C_2H_2+2Br_2\to C_2H_2Br_4\\ \text{Theo PT: }x+2y=n_{Br_2}=0,26(2)\\ (1)(2)\to\begin{cases} x=0,14\\ y=0,06 \end{cases} \\ \to \begin{cases} \%V_{C_2H_4}=\dfrac{0,14}{0,2}.100\%=70\%\\ \%V_{C_2H_2}=100-70=30\% \end{cases} \end{array}\)
\(\begin{array} {l} n_{Br_2}=\dfrac{41,6}{160}=0,26(mol)\\ \text{Đặt }n_{C_2H_4}=x(mol);n_{C_2H_2}=y(mol)\\ \to x+y=\dfrac{4,48}{22,4}=0,2(1)\\ C_2H_4+Br_2\to C_2H_4Br_2\\ C_2H_2+2Br_2\to C_2H_2Br_4\\ \text{Theo PT: }x+2y=n_{Br_2}=0,26(2)\\ (1)(2)\to\begin{cases} x=0,14\\ y=0,06 \end{cases} \\ \to \begin{cases} \%V_{C_2H_4}=\dfrac{0,14}{0,2}.100\%=70\%\\ \%V_{C_2H_2}=100-70=30\% \end{cases} \end{array}\)
a,\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: x x
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}65x+56y=30,7\\x+y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%m_{Zn}=\dfrac{0,3.65.100\%}{30,7}=63,52\%;\%m_{Fe}=100\%-63,52\%=36,48\%\)
b,
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,3 0,6
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,2 0,4
nHCl = 0,6+0,4 = 1 (mol)
\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)\)
a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(1\right)\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=0,25.2=0,5\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,3\left(mol\right)\\n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,3.22,4}{8,96}.100\%=75\%\\\%V_{C_2H_2}=25\%\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{C_2H_4Br_2}=n_{C_2H_4}=0,3\left(mol\right)\\n_{C_2H_2Br_4}=n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{C_2H_4Br_2}=0,3.188=56,4\left(g\right)\\m_{C_2H_2Br_4}=0,1.346=34,6\left(g\right)\end{matrix}\right.\)