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.

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

a)

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

\(a)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ 2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O\\ CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ b) n_{Br_2} = \dfrac{8}{160}=0,05(mol)\\ \Rightarrow n_{C_2H_2}= \dfrac{1}{2}n_{Br_2}= 0,025(mol)\\ n_{CO_2} = n_{CH_4} + 2n_{C_2H_2} = n_{CaCO_3} = \dfrac{50}{100} = 0,5(mol)\\ \Rightarrow n_{CH_4} = 0,5 - 0,025.2 = 0,45(mol)\\ \Rightarrow m = 0,45.16 + 0,05.26 = 8,5(gam)\)

\(\%m_{CH_4} = \dfrac{0,45.16}{8,5}.100\% = 84,7\%\\ \%m_{C_2H_2} = 100\% - 84,7\% = 15,3\%\)

\(n_{hh}=6,72:22,4=0,3mol\\ C_2H_2+2Br_2->C_2H_2Br_4\\ C_2H_4+Br_2->C_2H_2Br_2\\ n_{Br_2}=0,4mol\\ n_{C_2H_2}=a;n_{C_2H_4}=b\\ a+b=0,3\\ 2a+b=0,4\\ a=0,2;b=0,1\\ \%V_{C_2H_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\ \%V_{C_2H_4}=33,33\%\)

1/2 hỗn hợp có 0,1 mol C2H2 và 0,05mol C2H4

\(BT.C:n_{CO_2}=2n_{C_2H_2}+2n_{C_2H_4}=0,3mol\\ n_{CaCO_3}=n_{CO_2}=0,3\\ m_{KT}=0,3.100=30g\)

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

\(n_{C_2H_2Br_4}=\dfrac{6,72}{22,4}=0,3mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

 0,3                         0,3      ( mol )

\(\%C_2H_2=\dfrac{0,3}{0,6}.100=50\%\)

\(\%CH_4=100\%-50\%=50\%\)

b.

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

 0,3       0,6                                       ( mol )

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

 0,3          0,75                                        ( mol )

\(V_{O_2}=\left(0,6+0,75\right).22,4=1,35.22,4=30,24l\)

\(Đặt:n_{CH_4}=a\left(mol\right),n_{C_2H_2}=b\left(mol\right)\)

\(n_{hh}=a+b=0.35\left(mol\right)\left(1\right)\)

\(BTC:\)

\(a+2b=0.6\)

\(a=1\)

\(b=0.25\)

\(\%CH_4=\dfrac{0.1}{0.35}\cdot100\%=28.57\%\)

\(\%C_2H_2=71.43\%\)

\(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)⇒ x + y = \(\dfrac{7,84}{22,4} = 0,35(mol)\)

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_2 + \dfrac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + H_2O\\ CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\)

Theo PTHH : x + 2y = \(\dfrac{60}{100} = 0,6(2)\)

Từ (1)(2) suy ra x = 0,1 ; y = 0,25

Vậy : 

\(\%V_{CH_4} = \dfrac{0,1}{0,35}.100\% = 28,57\%\\ \%V_{C_2H_2} = 100\% - 28,57\% = 71,43\%\)

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

PTHH: C2H4 + Br2 -> C2H4Br2

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

Vhh khí = 2,8/22,4 = 0,125 (mol)

%VC2H4 = 0,05/0,125 = 40%

%CH4 = 100% - 40% = 60%

b, nCH4 = 0,125 - 0,05 = 0,075 (mol)

PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O

Mol: 0,05 ---> 0,15

CH4 + 2O2 -> (t°) CO2 + 2H2O

Mol: 0,075 ---> 0,15

Vkk = (0,15 + 0,15) . 5 . 22,4 = 33,6 (l)

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

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}n_{CO_2\left(CH_4\right)}=x\\n_{CO_2\left(C_2H_4\right)}=2y\end{matrix}\right.\)

\(n_{CaCO_3}=\dfrac{m_{CaCO_3}}{M_{CaCO_3}}=\dfrac{10}{100}=0,1mol\)

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

                    x+2y        x+2y             ( mol )

Ta có:

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

\(\%CH_4=\dfrac{0,05}{0,075}.100=66,66\%\)

\(\%C_2H_4=100\%-66,66\%=33,34\%\)

\(m_{CH_4}=0,05.16=0,8g\)

\(m_{C_2H_4}=0,025.28=0,7g\)