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n Br2=\(\dfrac{32}{160}\)=0,2 mol
C2H2+2Br2->C2H2Br4
0,1------0,2 mol
=>%VC2H2=\(\dfrac{0,1.22,4}{5,6}\).100=40%
=>%VCH4=100-40=60%
=>n CH4=\(\dfrac{5,6-0,1.22,4}{22,4}\)=0,15 mol
CH4+2O2-to>CO2+2H2O
0,15----0,3
C2H2+\(\dfrac{5}{2}\)O2-to>2CO2+H2O
0,1-----0,25 mol
=>VO2=(0,3+0,25).22,4=12,32l
a)
C2H4 + Br2 --> C2H4Br2
b) \(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,2<---0,2
=> \(\%V_{C_2H_4}=\dfrac{0,2.22,4}{8,96}.100\%=50\%\)
=> \(\%V_{CH_4}=100\%-50\%=50\%\)
\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{Br_2}=\dfrac{2,4}{160}=0,015mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
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.\(m_{dd.Br_2\left(tăng\right)}=m_{C_2H_2}=2,6g\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)
\(\%V_{C_2H_2}=\dfrac{0,1}{0,25}.100=40\%\)
\(\%V_{CH_4}=100\%-40\%=60\%\)
b.\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,1 0,2 ( mol )
\(C_{M\left(dd.Br_2\right)}=\dfrac{0,2}{0,1}=2M\)
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)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,5}.100\%=40\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,3.28}{0,3.28+0,2.26}.100\%=61,765\%\\\%m_{C_2H_2}=\dfrac{0,2.26}{0,3.28+0,2.26}.100\%=38,235\%\end{matrix}\right.\)
Câu 1:
Cho hỗn hợp qua bình Brom, chỉ C2H2 phản ứng
\(C_2H_2+Br_2\rightarrow C_2H_2Br_2\)
0,1______0,1________0,1
\(n_A=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{Br2}=0,5.0,2=0,1\left(mol\right)\)
\(n_{CH4}=0,2-0,1=0,1\left(mol\right)\)
a, \(\%V_{CH4}=\frac{0,1.22,4}{4,48}.100\%=50\%\)
\(\%V_{C2H2}=100\%-50\%=50\%\)
b,\(CM_{C2H2Br2}=\frac{0,1}{0,2}=0,5M\)
Câu 2:
1) Cho hh qua khí Brom , chỉ C2H2 phản ứng
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,1_____0,2_________0,2
\(n_{hh}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Br2}=0,2\left(mol\right)\)
\(n_{CH4}=0,15-0,1=0,05\left(mol\right)\)
\(V_{CH4}=0,05.22,4=1,12\left(l\right)\)
\(V_{C2H2}=0,1.22,4=2,24\left(l\right)\)
2) \(CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\)
0,25_______________0,125______
\(n_{CaC2}=\frac{8}{64}=0,125\left(mol\right)\)
\(\Rightarrow H=\frac{0,1}{0,125}.100\%=80\%\)