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a)
$C_2H_2 + 2Br_2 \to C_2H_2Br_4$
b) $n_{Br_2} = \dfrac{40.20\%}{160} = 0,05(mol)$
Theo PTHH : $n_{C_2H_2Br_4} = n_{C_2H_2} = \dfrac{1}{2}n_{Br_2} = 0,025(mol)$
$m_{C_2H_2Br_4} = 0,025.346 = 8,65(gam)$
c) $\%V_{C_2H_2} = \dfrac{0,025}{0,5}.100\% = 5\%$
$\%V_{CH_4} = 100\% - 5\% = 95\%$
\(C_2H_2+2Br_2->C_2H_2Br_4\\ n_{hh}=\dfrac{3,36}{22,4}=0,15mol\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1mol\\ n_{C_2H_2}=0,05mol\\ n_{Br_2}=2.0,05=0,1mol\\ m_{Br_2}=0,1.160=16g\\ \%V_{CH_4}=\dfrac{0,1}{0,15}.100\%=66,67\%\\ \%V_{C_2H_2}=33,33\%\)
\(n_{Br_2}=\dfrac{4}{160}=0,025mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,025 0,025 ( mol )
\(V_{hh}=\dfrac{2,8}{22,4}=0,125mol\)
\(\%V_{C_2H_4}=\dfrac{0,025}{0,125}.100=20\%\)
\(\%V_{CH_4}=100\%-20\%=80\%\)
\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{Br_2}=\dfrac{2,4}{160}=0,015mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,0075 0,015 ( mol )
\(V_{C_2H_2}=0,0075.22,4=0,168l\)
\(V_{CH_4}=3,36-0,168=3,192l\)
\(\%V_{C_2H_2}=\dfrac{0,168}{3,36}.100=5\%\)
\(\%V_{CH_4}=100\%-5\%=95\%\)
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
C2H2+2Br2->C2H2Br4
0,05-----0,1
n Br2=\(\dfrac{16}{160}\)=0,1 mol
=>VC2H2=0,05.22,4=1,12l
CaC2+2H2O->Ca(OH)2+C2H2
0,05------------------------------0,05
=>m CaC2=0,05.64=3,2g
\(n_{Br_2}=\dfrac{16}{160}=0,1mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,05 0,1 ( mol )
\(V_{C_2H_2}=0,05.22,4=1,12l\)
\(CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\)
0,05 0,05 ( mol )
\(m_{CaC_2}=0,05.64=3,2g\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
a) Ta có: \(n_{C_2H_4}=\dfrac{9,1}{28}=0,325\left(mol\right)=n_{Br_2}\) \(\Rightarrow V_{Br_2}=\dfrac{0,325}{2}=0,1625\left(l\right)=162,5\left(ml\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{C_2H_4}=0,325\left(mol\right)\\n_{CH_4}=\dfrac{13,44}{22,4}-0,325=0,275\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{hh}=9,1+0,275\cdot16=13,5\left(g\right)\)
c) PTHH: \(CH_4+2O_2 \underrightarrow{t^o} CO_2+2H_2O\)
\(C_2H_4+3O_2 \underrightarrow{t^o} 2CO_2+ 2H_2O\)
Theo các PTHH: \(\Sigma n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=1,525\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,525\cdot22,4=34,16\left(l\right)\)
\(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05mol\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,05 0,05 0,05 ( mol )
\(m_{Br_2}=0,05.160=8g\)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,25}.100=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)
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)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,5}.100\%=40\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,3.28}{0,3.28+0,2.26}.100\%=61,765\%\\\%m_{C_2H_2}=\dfrac{0,2.26}{0,3.28+0,2.26}.100\%=38,235\%\end{matrix}\right.\)
\(A.C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ B.n_{Br_2}=\dfrac{8}{160}=0,05mol\\ n_{C_2H_2Br_4}=\dfrac{1}{2}n_{Br_2}=0,025mol\\ m_{C_2H_2Br_4}=0,025.346=8,65g\)