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\(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right);n_{hh}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,025<-0,125
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,025}{0,125}.100\%=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)
a)\(m_{tăng}=m_{Br_2}=m_{C_2H_2}=0,78g\Rightarrow n_{C_2H_2}=0,03mol\)
\(n_{hh}=\dfrac{11,2}{22,4}=0,5mol\)
\(\Rightarrow n_{CH_4}=0,5-0,03=0,47mol\)
b)\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,03 0,03
\(m_{C_2H_2Br_4}=0,03\cdot266=7,98g\)
c)\(\%V_{C_2H_2}=\dfrac{0,03}{0,5}\cdot100\%=6\%\)
\(\%V_{CH_4}=100\%-6\%=94\%\)
Đặt `n_{CH_4}=x(mol);n_{C_2H_4}=y(mol);n_{H_2}=z(mol)`
`->x+y+z={3,36}/{22,4}=0,15(1)`
`C_2H_4+Br_2->C_2H_4Br_2`
`->m_{C_2H_4}=m_{\text{bình tăng}}=0,84(g)`
`->28y=0,84`
`->y=0,03(2)`
`M_A={0,975}/{{1,4}/{22,4}}=15,6(g//mol)`
`->m_A=15,6.0,15=2,34=16x+28y+2z(3)`
`(1)(2)(3)->x=0,09(mol);y=z=0,03(mol)`
Vậy trong hỗn hợp đầu:
`V_{H_2}=V_{C_2H_4}=0,03.22,4=0,672(l)`
`V_{CH_4}=0,09.22,4=2,016(l)`
\(a,n_{hh\left(CH_4,C_2H_4,C_2H_2\right)}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{hh\left(C_2H_4,C_2H_2\right)}=0,4-0,1=0,3\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a+b=0,3\\28a+26b=8,1\end{matrix}\right.\Leftrightarrow a=b=0,15\left(mol\right)\)
PTHH:
\(CH\equiv CH+2Br-Br\rightarrow CHBr_2-CHBr_2\)
\(CH_2=CH_2+Br-Br\rightarrow CH_2Br-CH_2Br\)
\(b,\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,4}.100\%=25\%\\\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,15}{0,4}.100\%=37,5\%\end{matrix}\right.\)
c, PTHH:
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\\ C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\\ \rightarrow n_{BaCO_3}=n_{CO_2}=0,1+0,15.0,15.2=0,7\left(mol\right)\\ m_{BaCO_3}=0,7.197=137,9\left(g\right)\)
a, C2H4 đã pư với dd Brom.
b, Ta có: \(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,05\left(mol\right)\Rightarrow m_{C_2H_4}=0,05.28=1,4\left(g\right)\)
a)
PTHH :
C2H4 + Br2 - > C2H4Br2
b) m( bình tăng ) = mC2H4 = 2,8(g) => nC2H4 = 0,1(mol)
nCH4 = nhh - nC2H4 = 0,15 - 0,1 = 0,05(mol)
=> mCH4 = 0,05.16 = 0,8(g)
c) Theo PTHH ta có : nBr2(pư)=nC2H4 = 0,1 (mol)
=> mBr2(pư) = 0,1.0 = 8(g)
Cảm ơn! ~~