a) m_tăng = m_C2H4 = 7 (g)

=> \(n_{C_2H_4}=\dfrac{7}{28}=0,25\left(mol\right)\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,25.24,79}{9,916}.100\%=62,5\%\\\%V_{CH_4}=100\%-62,5\%=37,5\%\end{matrix}\right.\)

b) 

PTHH: C₂H₄ + Br₂ ---> C₂H₄Br₂

           0,25-->0,25

=> \(C_{M\left(dd.Br_2\right)}=\dfrac{0,25}{0,25}=1M\)

c)

\(n_{CH_4}=\dfrac{9,916}{24,79}-0,25=0,15\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,15-->0,3

            C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

            0,25-->0,75

=> V_O2(đkc) = (0,3 + 0,75).24,79 = 26,0295 (l)

a) Khí thoát ra là CH₄

 \(n_{CH_4}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{3,7185}{14,874}.100\%=25\%\\\%V_{C_2H_4}=100\%-25\%=75\%\end{matrix}\right.\)

b) 

\(n_{C_2H_4}=\dfrac{14,874.75\%}{24,79}=0,45\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

            0,45-->0,45

=> \(C_{M\left(dd.Br_2\right)}=\dfrac{0,45}{0,15}=3M\)

c) 

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,15---------------------->0,3

            C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

             0,45------------------------->0,9

=> m_H2O = (0,3 + 0,9).18 = 21,6 (g)

a) Khi cho metan và axetilen qua dung dịch brom thì metan không phản ứng với brom nên thoát ra khỏi bình còn axetilen phản ứng với dung dịch brom.

=> 20,16 lít khí thoát là metan CH₄

=> V axetilen = 40,32 - 20,16 = 20,16 lít 

<=> %V _CH4  = %V C₂H₂ = 50%

b)

nCH₄ = nC₂H₂ = \(\dfrac{20,16}{22,4}\)= 0,9 lít

CH₄  +  2O₂ → CO₂  +  2H₂O

C₂H₂  +  \(\dfrac{5}{2}\)O₂   →  2CO₂  + H₂O

Theo tỉ lệ phản ứng cháy => nO₂ cần để đốt cháy hết hỗn hợp khí = 2nCH₄+\(\dfrac{5}{2}\)nC₂H₂= 4,05 mol.

=> V O₂ cần dùng = 4,05.22,4 = 90,72 lít

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ V_{CH_4}=4,48\left(l\right)\\ \Rightarrow V_{C_2H_2}=11,2-4,48=6,72\left(mol\right)\\ \Rightarrow\%V=\dfrac{6,72}{11,2}.100=60\%\)

\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4mol\)

\(m_{tăng}=m_{Br_2}=m_{C_2H_2}=2,6g\)

\(\Rightarrow n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

0,1          0,1

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

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

0,1         0,25      0,2

\(\Rightarrow n_{CO_2\left(CH_4\right)}=0,4-0,2=0,2mol\)

\(\Rightarrow n_{CH_4}=0,2mol\Rightarrow n_{O_2}=0,4mol\)

a)\(\%V_{CH_4}=\dfrac{0,2}{0,4}\cdot100\%=50\%\)

\(\%V_{C_2H_2}=100\%-50\%=50\%\)

b)\(\Sigma n_{O_2}=0,4+0,25=0,65mol\)

\(\Rightarrow V_{O_2}=0,65\cdot22,4=14,56l\)

\(\Rightarrow V_{kk}=14,56\cdot5=72,8l\)

a, \(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)

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

_____0,15____0,3 (mol)

\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,15.22,4}{11,2}.100\%=30\%\)

\(\Rightarrow\%V_{CH_4}=100-30=70\%\)

b, - Khí thoát ra ngoài là CH₄.

\(V_{CH_4}=11,2.70\%=7,84\left(l\right)\)

a)

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\)

b)

\(n_{CH_4} = n_{CO_2} = n_{CaCO_3} = \dfrac{20}{100} = 0,2(mol)\\ \%V_{CH_4} = \dfrac{0,2.22,4}{6,72}.100\% = 66,67\%\\ \%V_{C_2H_4} = 100\% -66,67\% = 33,33\%\)

a, \(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)

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

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

\(\Rightarrow V=V_{CH_4}=4,48-0,1.22,4=2,24\left(l\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{0,1.16+0,1.26}.100\%\approx38,1\%\\\%m_{C_2H_2}\approx61,9\%\end{matrix}\right.\)