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

\(n_{C2H4Br2}=n_{Br2}=n_{C2H4}=\frac{18,8}{188}=0,1\left(mol\right)\)

\(V\%_{C2H4}=\frac{0,1}{0,25}.100\%=40\%\)

\(V\%_{CH4}=100\%-40\%=60\%\)

Đổi 200ml = 0,2l

\(\Rightarrow CM_{Br2}=\frac{0,1}{0,2}=0,5M\)

Bài 14 : 

Vì metan không tác dụng với 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) \(n_{C2H4}=\dfrac{0,025.1}{1}=0,025\left(mol\right)\)

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

\(V_{CH4\left(dktc\right)}=1,4-0,56=0,84\left(l\right)\)

⁰/₀V_CH4 = \(\dfrac{0,84.100}{1,4}=60\)⁰/₀

⁰/₀V_C2H4 = \(\dfrac{0,56.100}{1,4}=40\)⁰/₀

 Chúc bạn học tốt

a) \(n_{C_2H_4Br_2}=\dfrac{4,7}{188}=0,025\left(mol\right)\)

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

           0,025<------------0,025

b) \(\left\{{}\begin{matrix}V_{C_2H_4}=0,025.22,4=0,56\left(l\right)\\V_{CH_4}=5,6-0,56=5,04\left(l\right)\end{matrix}\right.\)

a.\(m_{dd.Br_2\left(tăng\right)}=m_{C_2H_2}=2,6g\)

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

\(n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)

\(\%V_{C_2H_2}=\dfrac{0,1}{0,25}.100=40\%\)

\(\%V_{CH_4}=100\%-40\%=60\%\)

b.\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

    0,1         0,2                          ( mol )

\(C_{M\left(dd.Br_2\right)}=\dfrac{0,2}{0,1}=2M\)

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) \(n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

            0,1<----0,1<---0,1

=> \(m_{Br_2}=0,1.160=16\left(g\right)\)

b)

\(\%V_{C_2H_4}=\dfrac{0,1.22,4}{4}.100\%=56\%\)

=> \(\%V_{CH_4}=100\%-56\%=44\%\)

c) \(n_{CH_4}=\dfrac{4.44\%}{22,4}=\dfrac{11}{140}\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

          \(\dfrac{11}{140}\)-->\(\dfrac{11}{70}\)

            C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

           0,1---->0,3

=> \(V_{O_2}=\left(\dfrac{11}{70}+0,3\right).22,4=10,24\left(l\right)\)

=> V_kk = 10,24.5 = 51,2 (l)

1. \(n_{Br_2}=0,4.0,5=0,2\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)

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

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

Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,05\left(mol\right)\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,05.22,4}{1,4}.100\%=80\%\)

\(\Rightarrow\%V_{CH_4}=100-80=20\%\)

a)

C₂H₄ + Br₂ --> C₂H₄Br₂

b) \(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

              0,2<---0,2

=> \(\%V_{C_2H_4}=\dfrac{0,2.22,4}{8,96}.100\%=50\%\)

=> \(\%V_{CH_4}=100\%-50\%=50\%\)

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

\(\Rightarrow V_{CO_2}=\left(0,08+0,56\right).22,4=14,336\left(l\right)\)