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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 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
\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{Br_2}=\dfrac{2,4}{160}=0,015mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,0075 0,015 ( mol )
\(V_{C_2H_2}=0,0075.22,4=0,168l\)
\(V_{CH_4}=3,36-0,168=3,192l\)
\(\%V_{C_2H_2}=\dfrac{0,168}{3,36}.100=5\%\)
\(\%V_{CH_4}=100\%-5\%=95\%\)
\(C_2H_2+2Br_2->C_2H_2Br_4\\ n_{hh}=\dfrac{3,36}{22,4}=0,15mol\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1mol\\ n_{C_2H_2}=0,05mol\\ n_{Br_2}=2.0,05=0,1mol\\ m_{Br_2}=0,1.160=16g\\ \%V_{CH_4}=\dfrac{0,1}{0,15}.100\%=66,67\%\\ \%V_{C_2H_2}=33,33\%\)
a, \(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)
PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
_____0,15____0,3 (mol)
\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,15.22,4}{11,2}.100\%=30\%\)
\(\Rightarrow\%V_{CH_4}=100-30=70\%\)
b, - Khí thoát ra ngoài là CH4.
\(V_{CH_4}=11,2.70\%=7,84\left(l\right)\)
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
\(nC_2H_4=0,15\left(mol\right)\)
\(\Rightarrow nCH_3=0,35-0,15=0,2\left(mol\right)\)
\(\%VCH_4=\%nCH_4=\dfrac{0,2}{0,35}.100\%=57,14\%\)
\(\%VC_2H_4=100-57,14=42,86\%\)
a, PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
b, \(n_{Br_2}=\dfrac{5,6}{160}=0,035\left(mol\right)\)
Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,0175\left(mol\right)\)
\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,0175.22,4}{0,86}.100\%\approx45,58\%\)
\(\Rightarrow\%V_{CH_4}\approx54,42\%\)
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) $\%V_{C_2H_2} = \dfrac{0,025}{0,5}.100\% = 5\%$
$\%V_{CH_4} = 100\% - 5\% = 95\%$