\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{C_2H_4Br_2} = \dfrac{4,7}{199}= 0,025(mol)\\ n_{C_2H_6} = \dfrac{3-0,025.22,4}{22,4}= \dfrac{61}{560}(mol)\\ \Rightarrow m_{hh} = 0,025.28 + \dfrac{61}{560}30 \dfrac{1111}{280}\\ \Rightarrow \%m_{C_2H_4} = \dfrac{0,025.28}{ \dfrac{1111}{280}}.100\% = 17,64\%\\ \%m_{C_2H_6} = 100\% - 17,64\% = 82,86\%\)

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$C_2H_4 + Br_2 \to C_2H_4Br_2$

Ta có : 

$m_{C_2H_4} = m_{dd\ tăng} = 3(gam)$

$\Rightarrow n_{C_2H_4} = \dfrac{3}{28}(mol)$

$\Rightarrow n_{C_2H_6} = 0,25 - \dfrac{3}{28} = \dfrac{1}{7}(mol)$

$\%V_{C_2H_4} = \dfrac{ \dfrac{3}{28} }{0,25}.100\% = 42,9\%4

$\%V_{C_2H_6} = 100\% - 42,9\% = 57,1\%$

$\%m_{C_2H_4} = \dfrac{3}{3 + \dfrac{1}{7}.30}.100\% = 41,2\%$

$\%m_{C_2H_6} = 100\% - 41,2\% = 58,8\%$

$C_2H_4 + Br_2 \to C_2H_4Br_2$

Ta có : 

$m_{C_2H_4} = m_{dd\ tăng} = 3(gam)$

$\Rightarrow n_{C_2H_4} = \dfrac{3}{28}(mol)$

$\Rightarrow n_{C_2H_6} = 0,25 - \dfrac{3}{28} = \dfrac{1}{7}(mol)$

$\%V_{C_2H_4} = \dfrac{ \dfrac{3}{28} }{0,25}.100\% = 42,9\%$

$\%V_{C_2H_6} = 100\% - 42,9\% = 57,1\%$

$\%m_{C_2H_4} = \dfrac{3}{3 + \dfrac{1}{7}.30}.100\% = 41,2\%$

$\%m_{C_2H_6} = 100\% - 41,2\% = 58,8\%$

2c) tính thể tích O2 (đkc) để đốt hoàn toàn 1/4 hh X ban đầu

1,

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

b, \(n_{hh}=0,15\left(mol\right)\)

\(n_{C2H4}=n_{C2H4Br2}=\frac{4,7}{188}=0,025\left(mol\right)\)

\(V\%_{C2H4}=\frac{0,025.100}{0,15}=16,67\%\)

\(\Rightarrow\%V_{C2H6}=100\%-16,67\%=83,33\%\)

\(n_{C2H6}=0,15-0,025=0,125\left(mol\right)\)

\(\%m_{C2H4}=\frac{0,025.100.28}{0,025.28+0,125.30}=15,73\%\)

\(\Rightarrow\%m_{C2H6}=100\%-15,73\%=84,27\%\)

2,

a, \(C_3H_6+Br_2\rightarrow C_3H_6Br_2\)

\(m_{tang}=m_{C3H6}\Rightarrow n_{C3H6}=0,2\left(mol\right)\)

\(n_X=0,6\left(mol\right)\Rightarrow\%V_{C3H6}=33,33\%\)

\(\Rightarrow V\%_{C3H8}=100\%-33,33\%=66,67\%\)

b, \(n_{C3H8}=0,4\left(mol\right)\)

\(\overline{M_X}=\frac{m_{C3H6}+m_{C3H8}}{0,6}=\frac{130}{3}\)

\(d_{X/kk}=\frac{130}{87}\)

c, 1/4 X có 0,05 mol C3H6 ; 0,1 mol C3H8

\(C_3H_6+\frac{9}{2}O_2\underrightarrow{^{to}}3CO_2+3H_2O\)

\(C_3H_8+5O_2\underrightarrow{^{to}}3CO_2+4H_2O\)

\(\Rightarrow n_{O2}=\frac{9}{2}n_{C3H6}+5n_{C3H8}=0,725\left(mol\right)\)

\(\Rightarrow V_{O2}=0,725.22,4=16,24\left(l\right)\)

\(n_{hh}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(m_{Br_2}=200\cdot\dfrac{20}{100}=40\left(g\right)\)

\(n_{Br_2}=\dfrac{40}{160}=0.25\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(0.25........0.25..........0.25\)

\(\)\(n_{C_2H_4}=n_{hh}=0.25\left(mol\right)\)

=> Sai đề

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ m_{C_2H_4} = m_{tăng} = 10,5(gam)\\ n_{C_2H_6} = n_{hỗn\ hợp} - n_{C_2H_4} = \dfrac{10,08}{22,4} - \dfrac{10,5}{28} = 0,075(mol)\\ \Rightarrow m_{C_2H_6} = 0,075.30 = 2,25(gam)\)

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%

\(Đặt:n_{Al}=u\left(mol\right);n_{Fe}=v\left(mol\right)\left(u,v>0\right)\\ n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56u=22\\1,5a+u=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4\\u=0,2\end{matrix}\right.\\ \Rightarrow m_{Al}=0,4.27=10,8\left(g\right);m_{Fe}=56.0,2=11,2\left(g\right)\\ n_{HCl}=2.0,8=1,6\left(mol\right)\\ m_{HCl}=1,6.36,5=58,4\left(g\right)\\ m_{ddHCl}=\dfrac{58,4.100}{25}=233,6\left(g\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

x           3x             x             1,5x

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

y         2y              y          y

\(\left\{{}\begin{matrix}27x+56y=22\\1,5x+y=\dfrac{17,92}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

\(m_{Al}=0,4\cdot27=10,8g\)

\(m_{Fe}=22-10,8=11,2g\)

\(m_{HCl}=36,5\cdot\left(3x+2y\right)=36,5\cdot\left(3\cdot0,4+2\cdot0,2\right)=58,4g\)

\(m_{ddHCl}=\dfrac{m_{HCl}}{C\%}\cdot100\%=\dfrac{58,4}{25\%}\cdot100\%=233,6g\)