a)

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ b) n_{C_2H_4} = n_{Br_2} = \dfrac{8}{160}=0,05(mol)\\ n_{CaCO_3} = n_{CO_2} = n_{CH_4} + 2n_{C_2H_4} = \dfrac{50}{100} = 0,5(mol)\\ \Rightarrow n_{CH_4} = 0,5 - 0,05.2 = 0,4(mol)\\ \%m_{CH_4}= \dfrac{0,4.16}{0,4.16 + 0,05.28}.100\% = 82,05\%\\ \%m_{C_2H_4} =100\% - 82,05\% = 17,95\%\)

Sửa : 29,25 \(\to\) 29,55

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = \dfrac{8}{160} = 0,05(mol)\\ \Rightarrow m_{C_2H_4} = 0,05.28 = 1,4(gam)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 +3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ CO_2 + Ba(OH)_2 \to BaCO_3 + H_2O\\ n_{CO_2} = n_{CH_4} + 2n_{C_2H_4} = n_{CH_4} + 0,05.2 = n_{BaCO_3} = \dfrac{29,55}{197}=0,15(mol) \\ \Rightarrow n_{CH_4} = 0,05(mol)\\ \Rightarrow m_{CH_4} = 0,05.16 = 0,8(gam)\)

Đầu tiên, không có nước Br chỉ có nước Br2 em nhé!

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nBr2= 8/160=0,05(mol)

PTHH: C2H4 + Br2 -> C2H4Br2

nC2H4=nBr2=0,05(mol) => mC2H4=0,05.28=1,4(g)

- Khí bay ra là khí CH4.

CH4 + 2 O2 -to-> CO2 + 2 H2O

CO2 + Ba(OH)2 -> BaCO3 + H2O

nBaCO3=29,25/197= 117/ 788 (mol ) (Số xấu quá em ơi)

=> nCH4=nCO2=nBaCO3= 117/788(mol)

=> mCH4=16. 117/788= 468/197(g)

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

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

Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,05\left(mol\right)\)

\(\Rightarrow V_{C_2H_2}=0,05.22,4=1,12\left(l\right)\)

\(\Rightarrow V_{CH_4}=3,36-1,12=2,24\left(l\right)\)

b, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

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

\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,325\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,325.22,4=7,28\left(l\right)\Rightarrow V_{kk}=5V_{O_2}=36,4\left(l\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=0,2\left(mol\right)\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

\(\Rightarrow n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CaCO_3}=0,2.100=20\left(g\right)\)

a) CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

CO₂ + Ca(OH)₂ --> CaCO₃ + H₂O

b) \(n_{CH_4}=\dfrac{8}{16}=0,5\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

             0,5---->1-------->0,5

=> V_O2 = 1.22,4 = 22,4 (l)

c) 

PTHH: CO₂ + Ca(OH)₂ --> CaCO₃ + H₂O

             0,5------------------>0,5

=> m_CaCO3 = 0,5.100 = 50 (g)

\(n_{CH_4}=\dfrac{8}{18}=0,4\left(mol\right)\\ pthh:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\) 
           0,4      0,8      0,4 
=> \(v_{O_2}=0,8.22,4=17,92\left(l\right)\) 
\(CO_2+CaCO_3+H_2O\rightarrow Ca\left(HCO_3\right)_2\) 
0,4                                 0,4 
=> \(m_{Ca\left(HCO_3\right)_2}=0,4.214=85,6\left(g\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            FeS + 2HCl --> FeCl₂ + H₂S

=> \(n_{Fe}+n_{FeS}=n_{H_2}+n_{H_2S}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Và 56.n_Fe + 88.n_FeS = 18,8

=> \(\left\{{}\begin{matrix}n_{Fe}=0,1\left(mol\right)\\n_{FeS}=0,15\left(mol\right)\end{matrix}\right.\)

Bảo toàn S: n_CaSO3 = 0,15 (mol)

=> m = 0,15.120 = 18 (g)

=> B

B

nhh khí = 5,6/22,4 = 0,25 (mol)

nBr2 = 16/160 = 0,1 (mol)

PTHH: C2H2 + 2Br2 -> C2H2Br4

Mol: 0,05 <--- 0,1

nCH4 = 0,25 - 0,05 = 0,2 (mol)

%VC2H2 = 0,05/0,25 = 20%

%VCH4 = 100% - 20% = 80%

PTHH: 

2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O

0,05 ---> 0,125 ---> 0,1

CH4 + 2O2 -> (t°) CO2 + 2H2O

0,2 ---> 0,4 ---> 0,2

nCO2 = 0,2 + 0,1 = 0,3 (mol)

PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2O

nCaCO3 = 0,3 (mol)

mCaCO3 = 0,3 . 100 = 30 (g)

\(n_{hh}=\dfrac{11,2}{22,4}=0,5mol\)

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\)

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

  x                              x                     ( mol )

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

   y                              2y                   ( mol )

\(n_{CaCO_3}=\dfrac{80}{100}=0,8mol\)

\(Ca\left(OH\right)_2+CO_2\rightarrow\left(t^o\right)CaCO_3+H_2O\)

                     0,8                0,8                      ( mol )

Ta có:

\(\left\{{}\begin{matrix}x+y=0,5\\x+2y=0,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

\(\%V_{CH_4}=\dfrac{0,2}{0,5}.100=40\%\)

\(\%V_{C_2H_4}=100\%-40\%=60\%\)

\(m_{tăng}=m_{Ca\left(OH\right)_2}+m_{CaCO_3}=0,8.\left(74+100\right)=139,2g\)

Bài 7.

\(n_{CO_2}=\dfrac{1,344}{22,4}=0,06mol\Rightarrow n_C=0,06mol\Rightarrow m_C=0,72g\)

\(n_{H_2O}=\dfrac{1,62}{18}=0,09mol\Rightarrow n_H=0,18mol\Rightarrow m_H=0,18g\)

Ta có \(m_C+m_H=m_X\Rightarrow X\) chỉ chứa C và H.

Gọi CTHH là \(C_xH_y\)

\(x:y=\dfrac{m_C}{12}:\dfrac{m_H}{1}=\dfrac{0,72}{12}:\dfrac{0,18}{1}=0,06:0,18=1:3\)

\(\Rightarrow CH_3\)

Gọi CTPT là \(\left(CH_3\right)_n\Rightarrow M=15n\) (n∈N*)

Mà theo bài:

 \(22< M_X< 38\Rightarrow22< 15n< 38\Rightarrow1,467< n< 2,53\)

\(\Rightarrow n=2\Rightarrow C_2H_6\)

Chất X không làm mất màu dung dịch brom.

\(C_2H_6+Cl_2\underrightarrow{as}C_2H_5Cl+HCl\)

CTPT: C_nH_2n+2

\(n_{C_nH_{2n+2}}=\dfrac{2,9}{14n+2}\left(mol\right)\)

Bảo toàn C: \(n.\dfrac{2,9}{14n+2}=\dfrac{20}{100}=0,2\)

=> n = 4 

=> CTPT: C₄H₁₀

Phản ứng tách:

- Tách H₂: \(C_4H_{10}\underrightarrow{t^o,xt}C_4H_8+H_2\)

- Cracking: \(C_4H_{10}\underrightarrow{t^o,xt}\left[{}\begin{matrix}CH_4+C_3H_6\\C_2H_4+C_2H_6\end{matrix}\right.\)