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) C₂H₄ + Br₂ --> C₂H₄Br₂

C₂H₂ + 2Br₂ --> C₂H₂Br₄

b) Gọi số mol C₂H₄, C₂H₂ là a, b (mol)

=> a + b = \(\dfrac{1,68}{22,4}=0,075\left(mol\right)\) (1)

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

=> a + 2b = 0,1 (2)

(1)(2) => a = 0,05 (mol); b = 0,025 (mol)

=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,075}.100\%=66,67\%\\\%V_{C_2H_2}=\dfrac{0,025}{0,075}.100\%=33,33\%\end{matrix}\right.\)

c) 
PTHH: C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

           0,05--->0,15

             2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

            0,025-->0,0625

=> V_O2 = (0,15 + 0,0625).22,4 = 4,76 (l)

a.b.\(n_{Br_2}=\dfrac{16}{160}=0,1mol\)

\(n_{hh}=\dfrac{1,68}{22,4}=0,075mol\)

Gọi \(\left\{{}\begin{matrix}n_{C_2H_2}=x\\n_{C_2H_4}=y\end{matrix}\right.\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

  x          2x                            ( mol )

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

   y           y                           ( mol )

Ta có:

\(\left\{{}\begin{matrix}22,4x+22,4y=1,68\\2x+y=0,1\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x=0,025\\y=0,05\end{matrix}\right.\)

\(\%V_{C_2H_2}=\dfrac{0,025}{0,075}.100=33,33\%\)

\(\%V_{C_2H_4}=100\%-33,33\%=66,67\%\)

c.

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

 0,025    0,0625                                   ( mol )

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

0,05        0,15                                     ( mol )

\(V_{O_2}=\left(0,0625+0,15\right).22,4=4,76l\)

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

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

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, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

b, - Khí thoát ra là CH₄ ⇒ V_CH4 = 6,72 (l)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{6,72}{13,44}.100\%=50\%\\\%V_{C_2H_2}=50\%\end{matrix}\right.\)

a. PTHH: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

a. Vì CH₄ không phản ứng với dd Br₂ nên

       \(V_{CH_4}=6,72\left(l\right)\)

\(\%V_{CH_4}=\dfrac{6,72}{13,44}x100\%=50\%\)

\(\%V_{C_2H_2}=100\%-50\%=50\%\)

C2H4+Br2->C2H4Br2

0,05---0,05

n Br2=\(\dfrac{8}{160}\)=0,05 mol

V%C2H4=\(\dfrac{0,5.22,4}{22,4}\).100=5%

=>%CH4=95%

a) 
PTHH: C₂H₂+2Br₂ --> C₂H₂Br₄

b) \(n_{C_2H_2}=\dfrac{36}{26}=\dfrac{18}{13}\left(mol\right)\)

=> \(V_{C_2H_2}=\dfrac{18}{13}.22,4=\dfrac{2016}{65}\left(l\right)\)

\(n_{CH_4}=\dfrac{42-36}{16}=0,375\left(mol\right)\)

=> \(V_{CH_4}=0,375.22,4=8,4\left(l\right)\)

c) \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{\dfrac{2016}{65}}{\dfrac{2016}{65}+8,4}.100\%=78,69\%\\\%V_{CH_4}=\dfrac{8,4}{\dfrac{2016}{65}+8,4}.100\%=21,31\%\end{matrix}\right.\)

a, PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

b, \(n_{Br_2}=\dfrac{5,6}{160}=0,035\left(mol\right)\)

Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,0175\left(mol\right)\)

\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,0175.22,4}{0,86}.100\%\approx45,58\%\)

\(\Rightarrow\%V_{CH_4}\approx54,42\%\)

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 )

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

Ta có: \(n_{C_2H_2Br_2}=\dfrac{173}{186}\left(mol\right)\)

PT: \(C_2H_2+Br_2\rightarrow C_2H_2Br_2\)

Theo PT: \(n_{C_2H_2}=n_{C_2H_2Br_2}=\dfrac{173}{186}\left(mol\right)\)

\(\Rightarrow V_{C_2H_2}=\dfrac{173}{186}.22,4=20,83\left(l\right)\) > V_hh → vô lý

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