nhh = 5.6/22.4 = 0.25 (mol) 

nBr2 = 0.5*0.2 = 0.1 (mol) 

C2H4 + Br2 => C2H4Br2 

0.1_____0.1 

nCH4 = 0.25 - 0.1 = 0.15 (mol) 

VC2H4 = 0.1*22.4 = 2.24 (l) 

VCH4 = 0.15*22.4 = 3.36 (l) 

mCH4 = 0.15*16 = 2.4 (g) 

mC2H4 = 0.1*28 = 2.8 (g) 

%mCH4 = 2.4/(2.4 + 2.8) * 100% = 46.15%

%mC2H4 = 100 - 46.15 = 53.85%

Chúc bạn học tốt !!!

a) \(n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

            0,1<----0,1<---0,1

=> \(m_{Br_2}=0,1.160=16\left(g\right)\)

b)

\(\%V_{C_2H_4}=\dfrac{0,1.22,4}{4}.100\%=56\%\)

=> \(\%V_{CH_4}=100\%-56\%=44\%\)

c) \(n_{CH_4}=\dfrac{4.44\%}{22,4}=\dfrac{11}{140}\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

          \(\dfrac{11}{140}\)-->\(\dfrac{11}{70}\)

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

           0,1---->0,3

=> \(V_{O_2}=\left(\dfrac{11}{70}+0,3\right).22,4=10,24\left(l\right)\)

=> V_kk = 10,24.5 = 51,2 (l)

a, nC2H4Br2 = 28,2/188 = 0,15 (mol)

PTHH: C2H4 + Br2 -> C2H4Br2

Mol: 0,15 <--- 0,15 <--- 0,15

mC2H4 = 0,15 . 28 = 4,2 (g)

%mC2H4 = 4,2/9 = 46,67%

%mCH4 = 100% - 46,67% = 53,33%

b, VddBr2 = 0,15/0,5 = 0,3 (l)

c, PTHH: C2H4 + H2O -> (H2SO4 đặc) C2H5OH

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

nC2H5OH (TT) = 0,15 . 85% = 0,1275 (mol)

mC2H5OH (TT) = 0,1275 . 46 = 5,865 (g)

Cho hỗn hợp tác dụng với dung dịch Brom thu được sản phẩm là \(C_2H_4Br_2\).

\(n_{C_2H_4Br_2}=\dfrac{28,2}{188}=0,15mol\)

\(\Rightarrow n_{etilen}=0,15mol\)

\(\Rightarrow n_{metan}=\dfrac{9-0,15\cdot28}{16}=0,3mol\)

a)\(\%m_{CH_4}=\dfrac{0,3\cdot16}{9}\cdot100\%=53,33\%\)

\(\%m_{C_2H_4}=100\%-53,33\%=46,67\%\)

b)\(n_{Br_2}=n_{etilen}=0,15mol\)

\(\Rightarrow V_{ddBr_2}=\dfrac{n_{Br_2}}{C_{M_{Br_2}}}=\dfrac{0,15}{0,5}=0,3l\)

c)\(C_2H_4+H_2O\underrightarrow{xtH_2SO_4}C_2H_5OH\)

   0,15                              0,15

\(H=85\%\Rightarrow n_{etilen}=85\%\cdot0,15=0,1275mol\)

\(\Rightarrow n_{C_2H_5OH}=n_{C_2H_4}=0,1275mol\)

\(\Rightarrow m_{C_2H_5OH}=0,1275\cdot46=5,865g\)

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

Ta có: \(n_{Br_2}=\dfrac{5,6}{160}=0,035\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,035\left(mol\right)\)

\(\Rightarrow m_{C_2H_4}=0,035.28=0,98\left(g\right)\)

\(\Rightarrow m_{CH_4}=3,92-0,98=2,94\left(g\right)\)

b, Có: \(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{2,94}{3,92}.100\%=75\%\\\%m_{C_2H_4}=25\%\end{matrix}\right.\)

c, PT: \(C_2H_4+H_2O\xrightarrow[H^+]{t^o}C_2H_5OH\)

Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{C_2H_4}=0,035\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(LT\right)}=0,035.46=1,61\left(g\right)\)

Mà: \(m_{C_2H_5OH\left(TT\right)}=0,483\left(g\right)\)

\(\Rightarrow H\%=\dfrac{0,483}{1,61}.100\%=30\%\)

Bạn tham khảo nhé!

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

a) Ta có: \(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right)=n_{C_2H_4}\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,025}{\dfrac{5,6}{22,4}}\cdot100\%=10\%\) \(\Rightarrow\%V_{CH_4}=90\%\)

b) Theo PTHH: \(n_{C_2H_4Br_2}=n_{Br_2}=0,025mol\)

\(\Rightarrow m_{C_2H_4Br_2}=0,025\cdot188=4,7\left(g\right)\)

c) Ta có: \(n_{CH_4}=\dfrac{1}{22,4}=\dfrac{5}{112}\left(mol\right)=n_{O_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_4}=\dfrac{5}{112}\cdot16\approx0,71\left(g\right)\\m_{O_2}=\dfrac{5}{112}\cdot32\approx1,43\left(g\right)\end{matrix}\right.\)

  Vậy 1 lít Metan nhẹ hơn 1 lít Oxi

a)

\(m_{C_2H_2} = m_{tăng} = 5,2\ gam\\ \Rightarrow n_{C_2H_2} = \dfrac{5,2}{26} = 0,2(mol)\)

Vậy :

\(\%V_{C_2H_2} = \dfrac{0,2.22,4}{8,96}.100\% = 50\%\\ \%V_{CH_4} = 100\%-50\% = 50\%\)

b)

\(n_{CH_4} = n_{C_2H_2} = 0,2(mol)\)

CH₄  +  O₂   \(\xrightarrow{t^o}\)     CO₂      +    H₂O

0,2.........................0,2...................................(mol)

C₂H₂    +    \(\dfrac{5}{2}\)O₂ \(\xrightarrow{t^o}\)   2CO₂   +      H₂O

0,2................................0,4.................................(mol)

CO₂       +      Ca(OH)₂ → CaCO₃      +        H₂O

(0,2+0,4)............................(0,2+0,4)........................................(mol)

\(\Rightarrow m_{CaCO_3} =(0,2 + 0,4).100 = 60(gam)\)

a) \(n_{Br_2}=0,1.2=0,2\left(mol\right)\)

PTHH: \(CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

                  0,2<-------0,2

\(\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)

b) PTHH: \(CH\equiv CH+2Br_2\rightarrow CHBr_2-CHBr_2\)

                   0,1<--------0,2

\(\Rightarrow n_{C_2H_4}-n_{C_2H_2}=0,1\left(mol\right)\)