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\(CH_4+Cl_2\xrightarrow[]{a/s}CH_3Cl+HCl\\ CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\\ CH\equiv CH+2Br_2\rightarrow CHBr_2-CHBr_2\\ CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\\ C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\\ 2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\\ nCH_2=CH_2\xrightarrow[]{t^o,p,xt}\left(-CH_2-CH_2-\right)_n\)
Gọi n CH4 = a(mol) ; n C2H4 = b(mol) ; n C2H2 = c(mol)
=> 16a + 28b + 26c = 4,3(1)
$C_2H_4 + Br_2 \to C_2H_4Br_2$
$C_2H_2 + 2Br_2 \to C_2H_2Br_4$
n Br2 = b + 2c = 0,15(2)
Mặt khác :
m H2O = m tăng = 12,6 gam
=> n H2O = 0,7(mol)
n X = 8,96/22,4 = 0,4(mol)
Bảo toàn nguyên tố với H :
n H2O = 2n CH4 + 2n C2H4 + n C2H2
Ta có :
\(\dfrac{n_X}{n_{H_2O}} = \dfrac{a + b + c}{2a + 2b + c} = \dfrac{0,4}{0,7}(3)\)
Từ (1)(2)(3) suy ra a = 0,1 ; b = 0,05 ; c = 0,05
%V CH4 = 0,1/(0,1 + 0,05 + 0,05) .100% = 50%
%V C2H4 = %V C2H2 = 0,05/(0,1 + 0,05 + 0,05) .100% = 25%
\(n_{Br_2}=\dfrac{28}{160}=0,175\left(mol\right)\\ a,C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ b,n_{C_2H_4}=n_{Br_2}=0,175\left(mol\right)\\ V_{C_2H_4}=0,175.22,4=3,92\left(l\right)\\ \%V_{C_2H_4}=\dfrac{3,92}{8,6}.100\approx45,581\%\\ \%V_{CH_4}\approx54.419\%\)
c, Thiếu dữ kiện
\(n_{Br_2}=\dfrac{2,8}{160}=0,0175\left(mol\right)\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4}=n_{Br_2}=0,0175\left(mol\right)\\ m_{C_2H_4}=28.0,0175=0,49\left(g\right)\\ m_{CH_4}=4,3-0,49=3,81\left(g\right)\\ n_{CH_4}=\dfrac{3,81}{16}=0,238125\left(mol\right)\\ \%V_{C_2H_4}=\dfrac{0,0175}{0,0175+0,238125}.100\%\approx6,846\%\\ \%V_{CH_4}\approx93,154\%\)
Câu c hình như chưa đủ đề em hi
\(n_{hh}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{C_2H_4Br_2}=\dfrac{16}{188}=\dfrac{4}{47}mol\)
\(\Rightarrow n_{etilen}=\dfrac{4}{47}mol\)
\(\Rightarrow n_{metan}=0,3-\dfrac{4}{47}=\dfrac{101}{470}mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(\%V_{etilen}=\dfrac{\dfrac{4}{47}}{0,3}\cdot100\%=28,37\%\)
\(\%V_{metan}=100\%-28,37\%=71,63\%\)
a) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{6,4}{160}=0,04\left(mol\right);n_{hh}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,04<--0,04
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,04}{0,6}.100\%=6,67\%\\\%V_{CH_4}=100\%-6,67\%=93,33\%\end{matrix}\right.\)
b) \(n_{CH_4}=0,6-0,04=0,56\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,56----------->0,56
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)
0,04----------->0,08
\(\Rightarrow V_{CO_2}=\left(0,08+0,56\right).22,4=14,336\left(l\right)\)
Trùng hợp etilen