`a)` Khí etilen đã phản ứng với dung dịch brom.

`b)n_[Br_2]=24/160=0,15(mol)`

`C_2 H_4 +Br_2 ->C_2 H_4 Br_2`

 `0,15`      `0,15`                                   `(mol)`

    `=>m_[C_2 H_4]=0,15.28=4,2(g)`

a, C₂H₄ đã pư với dd Brom.

b, Ta có: \(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,05\left(mol\right)\Rightarrow m_{C_2H_4}=0,05.28=1,4\left(g\right)\)

a, Khí tác dụng với dd Brom: C₂H₄.

b, Ta có: \(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,05\left(mol\right)\Rightarrow m_{C_2H_4}=0,05.28=1,4\left(g\right)\)

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

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

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

0,05       0,05         0,05           ( mol )

\(m_{Br_2}=0,05.160=8g\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,25}.100=20\%\\\%V_{CH_4}=100\%-20\%=80\%\end{matrix}\right.\)

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)

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

ta có :

nBr2=\(\dfrac{16}{160}=0,1mol\)

C2H4+Br2->C2H4Br2

0,1------0,1

=>VC2H4=0,1.22,4=2,24l

=>VCH4=3,36l->n CH4=0,15 mol

->%VC2H4=\(\dfrac{2,24}{5,6}.100\)=40%

=>%VCH4=60%

c)

CH4+2O2-to>CO2+2H2O

0,15---------------0,15

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

0,1--------------------0,2
=>m CaCO3=0,35.100=35g

`a)` Khí axetilen đã phản ứng với dung dịch brom.

`b)n_[Br_2]=16/160=0,1(mol)`

`C_2 H_2 + 2Br_2 -> C_2 H_2 Br_4`

   `0,05`       `0,1`                                  `(mol)`

   `m_[C_2 H_2]=0,05.26=1,3(g)`

Ai giúp e với ạ