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

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

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

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

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

Giả sử các khí đo ở điều kiện sao cho 1 mol khi có thể tích V lít

\(n_{CO}+n_{CO_2}=\dfrac{20}{V}\)

\(n_{O_2}=\dfrac{8}{V}\left(mol\right)\)

PTHH: 2CO + O₂ --t^o--> 2CO₂

_____\(\dfrac{16}{V}\)<---\(\dfrac{8}{V}\)

=> \(\left\{{}\begin{matrix}n_{CO}=\dfrac{16}{V}\\n_{CO_2}=\dfrac{20}{V}-\dfrac{16}{V}=\dfrac{4}{V}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%V_{CO}=\dfrac{\dfrac{16}{V}}{\dfrac{20}{V}}.100\%=80\%\\\%V_{CO_2}=\dfrac{\dfrac{4}{V}}{\dfrac{20}{V}}.100\%=20\%\end{matrix}\right.\)

\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)

\(CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(C_2H_2+\dfrac{5}{2}O_2\rightarrow2CO_2+H_2O\)

Từ hai pt trên:\(\Rightarrow\left\{{}\begin{matrix}x+y=0,15\\x+2y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\%V_{CH_4}=\dfrac{0,1}{0,1+0,05}\cdot100\%=66,67\%\)

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

\(n_{CO_2}=\dfrac{V_{CO_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)

Gọi \(n_{CH_4}\) là x \(\Rightarrow V_{CH_4}=22,4x\)

      \(n_{C_2H_2}\) là y \(\Rightarrow V_{C_2H_2}=22,4y\)

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

  x                              x             ( mol )

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

    y                              2y                    ( mol )

Ta có:

\(\left\{{}\begin{matrix}22,4x+22,4y=3,36\\x+2y=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\Rightarrow V_{CH_4}=22,4.0,1=2,24l\)

\(\Rightarrow V_{C_2H_2}=22,4.0,05=1,12l\)

\(\%V_{CH_4}=\dfrac{2,24}{3,36}.100=66,67\%\)

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

nhh khí = 3,36/22,4 = 0,15 (mol)

Gọi nCH4 = a (mol); nC2H2 = b (mol)

a + b = 0,15 (1)

nCO2 = 4,48/22,4 = 0,2 (mol)

PTHH:

CH4 + 2O2 -> (t°) CO2 + 2H2O

a ---> 2a ---> a ---> 2a

2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O

b ---> 2,5b ---> 2b ---> b

a + 2b = 0,4 (2)

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

%VCH4 = 0,1/0,15 = 66,66%

%C2H2 = 100% - 66,66% = 33,34%

a)

\(\left\{{}\begin{matrix}V_{C_2H_2}=x\left(ml\right)\\V_{C_2H_4}=y\left(ml\right)\end{matrix}\right.\)⇒ x + y = 50(1)

\(C_2H_2 +\dfrac{5}{2} O_2 \xrightarrow{t^o} 2CO_2 + H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\)

Theo PTHH : 2,5x + 3y = 140(2)

Từ (1)(2) suy ra: x = 20 ; y = 30

Vậy :

\(\%V_{C_2H_2} = \dfrac{20}{50}.100\% = 40\%\\ \%V_{C_2H_4} = 100\% - 40\% = 60\%\)

b)

\(V_{CO_2} = 2V_{C_2H_2} + 2V_{C_2H_4} = 2.50 = 100(ml)\)