Ta có :

\(C_2H_4 + Br_2 \to C_2H_4Br_2 \\m_{C_2H_4} = m_{tăng} = 0,7(gam)\\ \Rightarrow V_{C_2H_4} = \dfrac{0,7}{28}.22,4 = 0,56(lít)\\ \Rightarrow V_{CH_4 } = 1,68 - 0,56 = 1,12(lít)\)

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

\(16n_{CH_4}+28n_{C_2H_4}=3\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%\approx53,33\%\\\%m_{C_2H_4}\approx46,67\%\end{matrix}\right.\)

- Ở cùng điều kiện nhiệt độ và áp suất, % số mol cũng là %V.

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)

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

Có: m _tăng = m_C2H4 = 0,05.28 = 1,4 (g)

a) \(n_{hh}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=0,15\\16a+28b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%=66,67\%\\\%V_{C_2H_4}=100\%-66,67\%=33,33\%\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%=53,33\%\\\%m_{C_2H_4}=100\%-53,33\%=46,67\%\end{matrix}\right.\)

b) \(m=m_{C_2H_4}=0,05.28=1,4\left(g\right)\)

\(m_{tăng}=m_{C_2H_2}=1,3\left(g\right)\\ \Rightarrow n_{C_2H_2}=\dfrac{1,3}{26}=0,05\left(mol\right)\\ \Rightarrow\%V_{\dfrac{C_2H_2}{A}}=\dfrac{0,05.22,4}{4,48}.100=25\%\\ \Rightarrow\%V_{\dfrac{CH_4}{A}}=100\%-25\%=75\%\)

Theo gt ta có: $n_{hh}=0,08(mol);n_{Br_2}=0,08(mol)$

$C_2H_2+2Br_2\rightarrow C_2H_2Br_4$

Suy ra $n_{C_2H_2}=0,04(mol)=n_{CH_4}$

a, $\Rightarrow \%V_{C_2H_2}=\%V_{C_2H_4}=50\%$

b, $CH_4+2O_2\rightarrow CO_2+2H_2O$

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

Ta có: $n_{O_2}=0,04.2+0,04.5=0,28(mol)\Rightarrow m_{O_2}=8,96(g)$

\(a)C_2H_2 +2Br_2 \to C_2H_2Br_2\\ n_{C_2H_2} = \dfrac{1}{2}n_{Br_2} = \dfrac{0,4.0,2}{2} = 0,04(mol)\\ \Rightarrow V_{C_2H_2} = 0,04.22,4 = 0,896(lít)\\ \%V_{C_2H_2} =\dfrac{0,896}{1,792}.100\% = 50\%\\ \Rightarrow \%V_{CH_4} = 100\% -50\% = 50\%\\ b)\\V_{CH_4} = V_{C_2H_2} = 0,896(lít)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_2 + \dfrac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + H_2O\\ \)

\(V_{O_2} = 2V_{CH_4} + \dfrac{5}{2}V_{C_2H_2} = 4,032(lít)\\ \Rightarrow m_{O_2} = \dfrac{4,032}{22,4}.32 = 5,76(gam)\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

Ta có: \(n_{CO_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)=n_{CH_4}\)

Đặt \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow a+b=\dfrac{5,04}{22,4}-0,075=0,15\)  (1)

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

            \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Theo PTHH: \(28a+26b=4,1\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{C_2H_4}=0,1\left(mol\right)\\b=n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)

Mặt khác: \(n_{hh}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,075}{0,225}\cdot100\%\approx33,33\%\\\%V_{C_2H_4}=\dfrac{0,1}{0,225}\cdot100\%\approx44,44\%\\\%V_{C_2H_2}=22,23\%\end{matrix}\right.\)

n_Br2 = 0,125 mol

Khi đốt cháy n_A = 0,25 mol

m_{NaOH ban đầu} = 36g => n_NaOH = 0,9 mol

Gọi x, y lần lượt là số mol của CO₂ và H₂O

m_dd = 180 + 44x + 18y

Vì NaOH dư do đó chỉ tạo muối trung hòa

CO₂ +2NaOH → Na₂CO₃ + H₂O

x          2x

n_{NaOH dư} = 0,9 – 2x

có 2 , 75 % = 40 ( 0 , 9 - 2 x ) 180 + 44 x + 18 y . 100 %  (1)

=> 81,21x + 0,495y = 31,05n

2,8 lít khí A tác dụng với 0,125 mol Br₂

=> 5,6 lít khí A tác dụng với 0,25 mol Br₂

Gọi số mol khí của CH₄, C₂H₄ và C₂H₂ lần lượt là a; b; c

Ta có a + b + c = 0,25 mol

Và b + 2c = 0,25

=> a = c

=> khi đốt cháy hỗn hợp A cho n_CO2 = n_H2O

Thay vào (1) => x = y = 0,38 mol

Bảo toàn C, H khi đốt cháy ta có

=> %V_CH4 = %V_C2H2 = 48%

%V_C2H4 = 4%

a) 
PTHH: C₂H₂+2Br₂ --> C₂H₂Br₄

b) \(n_{C_2H_2}=\dfrac{36}{26}=\dfrac{18}{13}\left(mol\right)\)

=> \(V_{C_2H_2}=\dfrac{18}{13}.22,4=\dfrac{2016}{65}\left(l\right)\)

\(n_{CH_4}=\dfrac{42-36}{16}=0,375\left(mol\right)\)

=> \(V_{CH_4}=0,375.22,4=8,4\left(l\right)\)

c) \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{\dfrac{2016}{65}}{\dfrac{2016}{65}+8,4}.100\%=78,69\%\\\%V_{CH_4}=\dfrac{8,4}{\dfrac{2016}{65}+8,4}.100\%=21,31\%\end{matrix}\right.\)

B5:

1)

\(V_{thoát}=V_{CH_4\left(đktc\right)}=3,36\left(l\right)\\ \Rightarrow\%V_{\dfrac{CH_4}{B}}=\dfrac{3,36}{4,48}.100=75\%\Rightarrow\%V_{\dfrac{C_2H_4}{B}}=100\%-75\%=25\%\)

2)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{CH_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{CaCO_3}=n_{CO_2}=n_{CH_4}=0,15\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{CaCO_3}=100.0,15=15\left(g\right)\)