a) C₂H₄ + Br₂ --> C₂H₄Br₂

b) \(n_{Br_2}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

           0,15<--0,15----->0,15

=> \(\%V_{C_2H_4}=\dfrac{0,15.22,4}{7,84}.100\%=42,857\%\)

=> \(\%V_{CH_4}=\dfrac{7,84-0,15.22,4}{7,84}.100\%=57,143\%\)

c) m_C2H4Br2 = 0,15.188 = 28,2 (g)

a, \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

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

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{72}{160}=0,45\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,15.28}{0,15.28+0,15.26}.100\%\approx51,85\%\\\%m_{C_2H_2}\approx48,15\%\end{matrix}\right.\)

\(\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,15.22,4}{6,72}.100\%=50\%\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{C_2H_4Br_2}=n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2Br_4}=n_{C_2H_2}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{C_2H_4Br_2}=0,15.188=28,2\left(g\right)\\m_{C_2H_2Br_4}=0,15.346=51,9\left(g\right)\end{matrix}\right.\)

\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{16}{160}=0,1mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

0,1         0,1                        ( mol )

\(\%V_{C_2H_4}=\dfrac{0,1.22,4}{16,8}.100=13,33\%\)

\(\%V_{CH_4}=100\%-13,33\%=86,67\%\)

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)

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

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%

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) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{6,4}{160}=0,04\left(mol\right);n_{hh}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

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

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