$C_2H_4 + Br_2 \to C_2H_4Br_2$

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 : 

$m_{C_2H_4} = m_{dd\ tăng} = 3(gam)$

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

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

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

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

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

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

a) \(V_{CH_4}=0,6\left(l\right)\)

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,6}{1,2}.100\%=50\%\\\%V_{C_2H_4}=100\%-50\%=50\%\end{matrix}\right.\)

b) \(n_{C_2H_4}=\dfrac{1,2-0,6}{24}=0,025\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂ 

         0,025-->0,025

=> \(m_{Br_2}=0,025.160=4\left(g\right)\)

c) 

\(n_{CH_4}=\dfrac{0,6}{24}=0,025\left(mol\right)\)

=> n_H = 0,025.4 = 0,1 (mol)

\(n_{Cl_2}=\dfrac{0,72}{24}=0,03\left(mol\right)\)

=> n_{Cl(thế H)} = 0,03 (mol)

Do n_H > n_{Cl(thế H)}

=> H không bị thế hoàn toàn bởi Cl

=> n_HCl = 0,03 (mol)

=> m_HCl = 0,03.36,5 = 1,095 (g)

$C_2H_4 + Br_2 \to C_2H_4Br_2$
Theo PTHH : 

$n_{C_2H_4} = n_{Br_2} = 0,1.2 = 0,2(mol)$
$\%m_{C_2H_4} = \dfrac{0,2.28}{10}.100\% = 56\%$

$\%m_{CH_4} = 100\% - 56\% = 44\%$

PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)

            \(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)

a) \(C_2H_4 + Br_2 \to C_2H_4Br_2\)

b)

\(n_{C_2H_4} = n_{Br_2} = \dfrac{5,6}{160} = 0,035(mol)\\ \%V_{C_2H_4} = \dfrac{0,035.22,4}{0,86}.100\% = 91,16\%\\ \%V_{CH_4} = 100\% - 91,16\% = 8,84\%\)

7,62 hay 6,72 :) ?

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

\(n_{Br_2}=\dfrac{32,4}{22,4}=\dfrac{81}{56}mol\Rightarrow n_{C_2H_2}=\dfrac{81}{112}mol\)

\(n_A=\dfrac{7,62}{22,4}=\dfrac{381}{1120}mol\)

\(\Rightarrow n_{CH_4}=\dfrac{381}{1120}-\dfrac{81}{112}=-\dfrac{429}{1120}\)

Đề vô lí