\(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05mol\)

\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

0,05       0,05         0,05           ( mol )

\(m_{Br_2}=0,05.160=8g\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,25}.100=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)

nC2H4Br2 = \(\dfrac{4,7}{188}\)=0,025(mol)

C2H4 + Br2 -> C2H4Br2

0,025 <-----------0,025

=>VC2H4 = 0,025 . 22,4=0,56(l)

=> VCH4 = 2,8 - 0,56 =2,24 (l)

%VCH4 =\(\dfrac{2,24.100}{2,8}\)=80%

%VC2H4 = 100 % -80% = 20%

Bài 9 : 

Metan không tác dụng với dung dịch Brom nên :

\(n_{C2H4Br2}=\dfrac{4,7}{188}=0,025\left(mol\right)\)

a) Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_2|\)

               1          1               1

            0,025                      0,025

b) \(n_{C2H4}=\dfrac{0,025.1}{1}=0,025\left(mol\right)\)

\(V_{C2H4\left(dktc\right)}=0,025.22,4=0,56\left(l\right)\)

\(\%V_{C2H4}=\dfrac{0,56.100}{2,8}=20\%\)

\(\%V_{CH4}=100\%-20\%=80\%\)

 Chúc bạn học tốt

a, - Khí pư với Brom là C₂H₄

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b, Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,1.28}{4}.100\%=70\%\\\%m_{CH_4}=100-70=30\%\end{matrix}\right.\)

C2H2+2Br2-to>C2H2Br4

0,4-------0,8----------0,4 mol

n C2H2Br4=\(\dfrac{138,4}{346}=0,4mol\)

=>m  Br2= 0,8.160=128g

=>m ddBr2=1280g

=>%mC2H2=\(\dfrac{0,4.22,4}{16,8}.100=53,3\%\)

=>%m CH4=100-53,3=46,7%

a) C₂H₄ + Br₂ --> C₂H₄Br₂

b) n_Br2 = 0,2.0,2 = 0,04 (mol)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

            0,04<--0,04

=> \(m_{C_2H_4}=0,04.28=1,12\left(g\right)\)

\(m_{CH_4}=n_{CH_4}.M_{CH_4}=\left(\dfrac{1,12}{22,4}-0,04\right).16=0,16\left(g\right)\)

c) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,04.22,4}{1,12}.100\%=80\%\\\%V_{CH_4}=100\%-80\%=20\%\end{matrix}\right.\)

a) PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b) Ta có: \(n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{C_2H_4}\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,07\cdot22,4}{1,72}\cdot100\%\approx91,16\%\) 

\(\Rightarrow\%V_{CH_4}=8,84\%\)

a) \(C_2H_4 + Br_2 \to C_2H_4Br_2\)

b)

\(n_{C_2H_4} = n_{Br_2} = \dfrac{5,6}{160} = 0,035(mol)\\ \%V_{C_2H_4} = \dfrac{0,035.22,4}{0,86}.100\% = 91,16\%\\ \%V_{CH_4} = 100\% - 91,16\% = 8,84\%\)

a.b.\(m_{tăng}=m_{C_2H_4}=2,8g\)

\(n_{C_2H_4}=\dfrac{2,8}{28}=0,1mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(n_{hh}=\dfrac{6,72}{22,4}=0,3mol\)

\(\rightarrow m_{CH_4}=\left(0,3-0,1\right).16=3,2g\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{2,8}{2,8+3,2}.100=46,67\%\\\%m_{CH_4}=100\%-46,67\%=53,33\%\end{matrix}\right.\)

c.\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

    0,2                            0,2                 ( mol )

\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)

  0,1                              0,2                ( mol )

\(V_{CO_2}=\left(0,2+0,2\right).22,4=8,96l\)

.