\(n_{C_2H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{t^0}}2CO_2+H_2O\)

\(0.4.........1\)

\(V_{O_2}=1\cdot22.4=22.4\left(l\right)\)

\(V_{kk}=5V_{O_2}=5\cdot22.4=112\left(l\right)\)

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 )

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

a)

C₂H₄ + Br₂ --> C₂H₄Br₂

C₂H₂ + 2Br₂ --> C₂H₂Br₄

b) 

Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\) => \(a+b=\dfrac{13,44}{22,4}=0,6\) (1)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

               a--->a

            C₂H₂ + 2Br₂ --> C₂H₂Br₄

               b--->2b

=> \(a+2b=0,8.1=0,8\) (2)

(1)(2) => a = 0,4 (mol); b = 0,2 (mol)

PTHH: C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

             0,4--->1,2

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

               0,2---->0,5

=> \(V_{O_2}=\left(1,2+0,5\right).22,4=38,08\left(l\right)\)

=> V_kk = 38,08 : 20% = 190,4 (l)

\(a) C_2H_4 + Br_2 \to C_2H_4Br_2\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ b) n_{C_2H_4} = a(mol) ; n_{C_2H_2} = b(mol)\\ n_X = a + b = \dfrac{0,56}{22,4} = 0,025(mol)\\ n_{Br_2} = a + 2b = \dfrac{5,6}{160} =0,035(mol)\\ \Rightarrow a = 0,015 ; b = 0,01\\ \%V_{C_2H_4} = \dfrac{0,015}{0,025}.100\% = 60\%\\ \%V_{C_2H_2} = 100\% -60\% = 40\%\)

\(c) C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ 2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O\\ n_{O_2} = 3n_{C_2H_4} + \dfrac{5}{2}n_{C_2H_2} = 0,07(mol)\\ V_{O_2} = 0,07.22,4 = 1,568(lít)\)

C2H2+2Br2->C2H2Br4

x-----------2x

C2H4+Br2->C2H4Br2

y-----------2y

n Br2=0,8 mol

\(\left\{{}\begin{matrix}x+y=0,6\\2x+y=0,8\end{matrix}\right.\)

=>x=0,2 ,y=0,4 mol

=>%VC2H2=\(\dfrac{0,2.22,4}{13,44}100\)=33,33%

=>%C2H4=66,67%

C2H4+3O2-tO>2CO2+2H2O

C2H2+5\2O2-to>2CO2+H2O

=>Vkk=1,7.22,4.5=190,4l

\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)

\(n_{Br_2}=0,4\cdot2=0,8mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

x             x            x

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

y             2y            y

\(\Rightarrow\left\{{}\begin{matrix}x+y=0,6\\x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

a)\(\%V_{C_2H_4}=\dfrac{0,4}{0,4+0,2}\cdot100\%=66,67\%\)

   \(\%V_{C_2H_2}=100\%-66,67\%=33,33\%\)

b)\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

   \(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\)

   \(\Rightarrow n_{O_2}=3n_{C_2H_4}+\dfrac{5}{2}n_{C_2H_2}=3\cdot0,4+\dfrac{5}{2}\cdot0,2=1,7mol\)

  \(\Rightarrow V_{O_2}=1,7\cdot22,4=38,08l\)

  \(\Rightarrow V_{kk}=5V_{O_2}=5\cdot38,08=190,4l\)

a, Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(2C_2H_2+5O_2\underrightarrow{^{t^o}}4CO_2+2H_2O\)

\(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,5\left(mol\right)\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)

b, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=56\left(l\right)\)

c, - Hiện tượng: Br₂ nhạt màu dần.

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

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{CO_2}=2.0,25=0,5\left(mol\right)\\ a,V_{CO_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{O_2}=\dfrac{5}{2}.0,25=0,625\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,625.22,4=14\left(l\right)\\ V_{kk\left(đkct\right)}=\dfrac{100}{20}.14=70\left(lít\right)\)