Gọi :n C2H4 = a(mol) ; n C3H6 = b(mol) ; n C2H6 = c(mol)

n X = a + b + c = 2,016/22,4 = 0,09(1)

Bảo toàn nguyên tố với C :

n CO2 = 2a + 3b + 2c = 4,704/22,4 = 0,21(2)

Mặt khác :

n Br2 = n C2H4 + n C3H6 = a + b = 9,6/160 = 0,06(3)

Từ (1)(2)(3) suy ra a = b = c = 0,03

Suy ra m X = 0,03.28 + 0,03.42 + 0,03.30 = 3(gam)

%m C2H4 = 0,03.28/3 .100% = 28%

%m C3H6 = 0,03.42/3 .100% = 42%

%m C2H6 = 100% -28% -42% = 30%

\(n_{Br_2}=0.3\cdot1=0.3\left(mol\right)\)

\(C_3H_6+Br_2\rightarrow C_3H_6Br_2\)

\(0.3.........0.3\)

\(n_{CO_2}=\dfrac{48.4}{44}=1.1\left(mol\right)\)

\(BảotoànC:\)

\(n_{C_2H_6}=\dfrac{1.1-0.3\cdot3}{2}=0.2\left(mol\right)\)

\(\%C_2H_6=\dfrac{0.2\cdot30}{0.2\cdot30+0.3\cdot42}\cdot100\%=32.25\%\)

\(\%C_3H_6=67.75\%\)

n CH4 = a(mol) ; n C2H6 = b(mol) ; n C2H4 = c(mol)

=> 16a + 30b + 28c = 11,8(1)

Bảo toàn nguyên tố với C :

n CO2 = a + 2b + 2c = 35,2/44 = 0,8(2)

Mặt khác : 

n X = 20,16/22,4 = 0,9(mol)

n Br2 = n C2H4 = 48/160 = 0,3(mol)

Suy ra : 

n X / n Br2 = (a + b + c)/c = 0,9/0,3(3)

Từ (1)(2) (3) => a = 0,1 ; b = 0,2 ; c = 0,15

Vậy :

%V CH4 = 0,1/(0,1 + 0,2 + 0,15)    .100% =22,22%
%V C2H6 = 0,2/(0,1 + 0,2 + 0,15)   .100% = 44,44%
%V C2H4 = 100% -22,22% -44,44% = 33,34%

\(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\)\(\Rightarrow a + b = \dfrac{3,36}{22,4} = 0,15(1) \)

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ n_{CO_2} = a + 2b = \dfrac{4,48}{22,4} = 0,2(2)\)

Từ (1)(2) suy ra: a = 0,1 ; b = 0,05

Suy ra:

\(\%V_{CH_4} = \dfrac{0,1}{0,15}.100\% = 66,67\%\\ \%V_{C_2H_4} = 100\% - 66,67\% = 33,33\%\)

b)

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{Br_2} = n_{C_2H_4} = 0,05(mol)\\ \Rightarrow m_{Br_2} = 0,05.160 = 8\ gam\)

\(n_{Br_2}=\dfrac{16}{160}=0,1mol\Rightarrow n_{CH_2}=0,1mol\)

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

\(\Rightarrow n_{CH_4}=0,25-0,1=0,15mol\)

\(\%V_{CH_2}=\dfrac{0,1}{0,25}\cdot100\%=40\%\)

\(\%V_{CH_4}=100\%-40\%=60\%\)

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

\(CH_2+\dfrac{3}{2}O_2\underrightarrow{t^o}CO_2+H_2O\)

\(\Rightarrow\Sigma n_{CO_2}=0,15+0,1=0,25mol\)

\(BTC:n_{CO_2}=n_{CaCO_3}=0,25mol\)

\(\Rightarrow m_{\downarrow}=0,25\cdot100=25g\)

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)

Gọi \(n_{CH_4}=a\left(mol\right)\) và \(n_{C_2H_6}=b\left(mol\right)\)

\(n_{hh}=\dfrac{11,1555}{22,4}=0,5\)

\(\Rightarrow a+b=0,5\left(1\right)\)

\(n_{CO_2}=\dfrac{16,1135}{22,4}=0,72mol\)

Bảo toàn C: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_6}=a+2b=0,72\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\left\{{}\begin{matrix}a=0,28mol\\b=0,22mol\end{matrix}\right.\)

\(\%V_{CH_4}=\dfrac{0,28}{0,28+0,22}\cdot100\%=56\%\)

\(\%V_{C_2H_6}=100\%-56\%=44\%\)