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\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{Br_2}=\dfrac{2,4}{160}=0,015mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,0075 0,015 ( mol )
\(V_{C_2H_2}=0,0075.22,4=0,168l\)
\(V_{CH_4}=3,36-0,168=3,192l\)
\(\%V_{C_2H_2}=\dfrac{0,168}{3,36}.100=5\%\)
\(\%V_{CH_4}=100\%-5\%=95\%\)
a)
C2H4 + Br2 --> C2H4Br2
C2H2 + 2Br2 --> C2H2Br4
b) Gọi số mol C2H4, C2H2 là a, b (mol)
=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (1)
PTHH: C2H4 + Br2 --> C2H4Br2
a---->a
C2H2 + 2Br2 --> C2H2Br4
b---->2b
=> a + 2b = \(\dfrac{112}{160}=0,7\) (2)
(1)(2) => a = 0,3 (mol); b = 0,2 (mol)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,5}.100\%=40\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,3.28}{0,3.28+0,2.26}.100\%=61,765\%\\\%m_{C_2H_2}=\dfrac{0,2.26}{0,3.28+0,2.26}.100\%=38,235\%\end{matrix}\right.\)
\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{16}{160}=0,1mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,1 0,1 ( mol )
\(\%V_{C_2H_4}=\dfrac{0,1.22,4}{16,8}.100=13,33\%\)
\(\%V_{CH_4}=100\%-13,33\%=86,67\%\)
a)
C2H4 + Br2 --> C2H4Br2
b) \(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,2<---0,2
=> \(\%V_{C_2H_4}=\dfrac{0,2.22,4}{8,96}.100\%=50\%\)
=> \(\%V_{CH_4}=100\%-50\%=50\%\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Gọi: \(\left\{{}\begin{matrix}n_{C_2H_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow x+y=\dfrac{2,24}{22,4}=0,1\left(mol\right)\left(1\right)\)
\(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=x+2y=\dfrac{24}{160}=0,15\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x=y=0,05\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,05.22,4}{2,24}.100\%=50\%\)
a) C2H4 + Br2 --> C2H4Br2
b) \(n_{Br_2}=\dfrac{24}{160}=0,15\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,15<--0,15----->0,15
=> \(\%V_{C_2H_4}=\dfrac{0,15.22,4}{7,84}.100\%=42,857\%\)
=> \(\%V_{CH_4}=\dfrac{7,84-0,15.22,4}{7,84}.100\%=57,143\%\)
c) mC2H4Br2 = 0,15.188 = 28,2 (g)
nC2H4Br2 = \(\dfrac{4,7}{188}\)=0,025(mol)
C2H4 + Br2 -> C2H4Br2
0,025 <-----------0,025
=>VC2H4 = 0,025 . 22,4=0,56(l)
=> VCH4 = 2,8 - 0,56 =2,24 (l)
%VCH4 =\(\dfrac{2,24.100}{2,8}\)=80%
%VC2H4 = 100 % -80% = 20%
Bài 9 :
Metan không tác dụng với dung dịch Brom nên :
\(n_{C2H4Br2}=\dfrac{4,7}{188}=0,025\left(mol\right)\)
a) Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_2|\)
1 1 1
0,025 0,025
b) \(n_{C2H4}=\dfrac{0,025.1}{1}=0,025\left(mol\right)\)
\(V_{C2H4\left(dktc\right)}=0,025.22,4=0,56\left(l\right)\)
\(\%V_{C2H4}=\dfrac{0,56.100}{2,8}=20\%\)
\(\%V_{CH4}=100\%-20\%=80\%\)
Chúc bạn học tốt
a, \(n_{hh}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
\(n_{Br2}=\dfrac{24}{160}=0,15\left(mol\right)\)
\(CH_2=CH_2+Br_2\rightarrow CH_2+CH_2\)
/Br /Br
0,15mol<----- 0,15mol
\(nC_2H_4=0,15\left(mol\right)\)
\(\Rightarrow nCH_3=0,35-0,15=0,2\left(mol\right)\)
\(\%VCH_4=\%nCH_4=\dfrac{0,2}{0,35}.100\%=57,14\%\)
\(\%VC_2H_4=100-57,14=42,86\%\)
á quên câu cuối bổ sung:
mC2H4 = 0,15 . 28 = 4,2 g