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

Đặt \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\)  ( mol ) \(\Rightarrow m_{hh}=16x+28y=6\left(g\right)\)   (1)

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

 x                             x                    ( mol )

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

  y                               2y                    ( mol )

\(n_{CO_2}=x+2y=0,4\left(mol\right)\)    (2)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\%m_{CH_4}=\dfrac{0,2.16}{6}.100=53,33\%\)

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

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

  0,1         0,1                   ( mol )

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

a, Ta có \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)

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

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,05\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,05.22,4=1,12\left(l\right)\\V_{C_2H_4}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)

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

c, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)

\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)

a)

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4Br_2} = n_{C_2H_4} = n_{Br_2} = \dfrac{160.5\%}{160} = 0,05(mol)\\ \Rightarrow m_{C_2H_4Br_2} = 0,05.188 = 9,4(gam)\)

b)

\(\%V_{C_2H_4} = \dfrac{0,05.22,4}{4,48}.100\% = 25\%\\ \%V_{CH_4} = 100\% - 25\% = 75\%\)

 Ta có Cùng điều kiện -> Quy số lít về số mol.n(hh ban đầu) = 20 mol; n(hh sau) = 16 lít

=> H2 phản ứng mất 4 lít => C2H2 có 2 lít và CH4 có 8 lít

a.\(n_{hh}=\dfrac{V_{hh}}{22,4}=\dfrac{0,896}{22,4}=0,04mol\)

\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{8}{160}=0,05mol\)

Gọi \(n_{C_2H_2}\) là x => \(V_{C_2H_2}=22,4x\)

       \(n_{C_2H_4}\) là y => \(V_{C_2H_4}=22,4y\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

  x           2x                 x         ( mol )

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

   y         y                y            ( mol )

Ta có hpt:

\(\left\{{}\begin{matrix}22,4x+22,4y=0,896\\2x+y=0,05\end{matrix}\right.\)   \(\Leftrightarrow\left\{{}\begin{matrix}x=0,01\\y=0,03\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{C_2H_2}=0,01.22,4=0,224\\V_{C_2H_4}=0,03.22,4=0,672\end{matrix}\right.\)

\(\%V_{C_2H_2}=\dfrac{0,224}{0,896}.100=25\%\)

\(\%V_{C_2H_4}=100-25=75\%\)

b.\(m_{Sp}=0,01.186+0,03.346+0,03.188=16,02g\)

( Em cảm ơn anh Kudo Sinichi rất nhiều ạ!! )

a) Gọi số mol CH₄, C₂H₄ là a, b (mol)

=> \(\left\{{}\begin{matrix}16a+28b=11,6\\a+b=\dfrac{11,2}{22,4}=0,5\end{matrix}\right.\)

=> a = 0,2 (mol); b = 0,3 (mol)

\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2}{0,5}.100\%=40\%\\\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,2.16}{11,6}.100\%=27,586\%\\\%m_{C_2H_4}=\dfrac{0,3.28}{11,6}.100\%=72,414\%\end{matrix}\right.\)

b) 

\(n_{C_2H_4}=\dfrac{5,6.60\%}{22,4}=0,15\left(mol\right)\)

m_tăng = m_C2H4 = 0,15.28 = 4,2 (g)

\(\left\{{}\begin{matrix}CH_4:x\left(mol\right)\\C_2H_2:y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}16x+26y=11,6\\x+y=\dfrac{11,2}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,14mol\\y=0,36mol\end{matrix}\right.\)

a)\(\%V_{CH_4}=\dfrac{0,14}{0,5}\cdot100\%=28\%\)

\(\%V_{C_2H_2}=100\%-28\%=72\%\)

\(\%m_{CH_4}=\dfrac{0,14\cdot16}{11,6}\cdot100\%=19,31\%\)

\(\%m_{C_2H_2}=100\%-19,31\%=80,69\%\)

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

Dẫn dung dịch qua bình đựng brom chỉ có \(C_2H_2\) tác dụng.

\(\Rightarrow m_{tăng}=m_{C_2H_2}=0,25\cdot26=6,5g\)

\(m_Q=\left(7,7.2\right).0,1=1,54\left(g\right)\)

=> m_T = 1,54 (g)

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{H_2}=c\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a+b+c=0,11\left(1\right)\\16a+28b+2c=1,54\left(2\right)\end{matrix}\right.\)

n_giảm = n_H2(pư) = 0,11 - 0,1 = 0,01 (mol)

\(n_{Br_2}=\dfrac{4,8}{160}=0,03\left(mol\right)\)

Bảo toàn liên kết: b = 0,01 + 0,03 = 0,04 (mol) (3)

(1)(2)(3) => a = 0,02 (mol); b = 0,04 (mol); c = 0,05 (mol)

=> n_H2(Q) = 0,05 - 0,01 = 0,04 (mol)

=> \(\%V_{H_2}=\dfrac{0,04}{0,1}.100\%=40\%\)

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

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

____0,15___0,3 (mol)

\(\Rightarrow\left\{{}\begin{matrix}V_{C_2H_2}=0,15.22,4=3,36\left(l\right)\\V_{CH_4}=2,24\left(l\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

$C_2H_2+2Br_2\to Br_2CH-CHBr_2$

$n_{Br_2}=\dfrac{48}{160}=0,3(mol)$

$\Rightarrow n_{C_2H_2}=\dfrac{n_{Br_2}}{2}=0,15(mol)$

$\Rightarrow V_{C_2H_2}=0,15.22,4=3,36(l)$

$\Rightarrow V_{CH_4}=5,6-3,36=2,24(l)$