Khí thoát ra là metan.

CTHH của A :C_nH_2n

Suy ra: 

\(n_A = \dfrac{5,6-2,688}{22,4} = 0,13(mol)\\ m_A =m_{tăng} = 3,64(gam)\)

Suy ra: 0,13.14n = 3,64 ⇒ n = 2

Vậy A là \(C_2H_2\)(Axetilen)

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

b, - Khí thoát ra là CH₄.

⇒ V_CH4 = 4,48 (l)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{4,48}{11,2}.100\%=40\%\\\%V_{C_2H_4}=100-40=60\%\end{matrix}\right.\)

a) 

PTHH: C₂H₂ + 2Br₂ --> C₂H₂Br₄

Khí thoát ra là CH₄

\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{C_2H_2}=\dfrac{6,72}{22,4}-0,2=0,1\left(mol\right)\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,2.16}{0,2.16+0,1.26}.100\%=55,17\%\\\%m_{C_2H_2}=\dfrac{0,1.26}{0,2.16+0,1.26}.100\%=44,83\%\end{matrix}\right.\)

b) 

PTHH: C₂H₂ + 2Br₂ --> C₂H₂Br₄

              0,1--->0,2

=> m_Br2 = 0,2.160 = 32 (g)

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

b) Ta có: \(n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{C_2H_4}\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,07\cdot22,4}{1,72}\cdot100\%\approx91,16\%\) 

\(\Rightarrow\%V_{CH_4}=8,84\%\)

Chọn D

n_X = 1 mol; M_X = 5,5.4 = 22

n_Z = 0,2 mol; M_Z = 11,5.2 = 23

BTKL: m _{bình Br2 tăng}  = m_X – m_Z = 1.22 – 0,2.23 = 17,4 gam

Đáp án D

n_X = 1 mol; M_X = 5,5.4 = 22

n_Z = 0,2 mol; M_Z = 11,5.2 = 23

BTKL: m _{bình Br2 tăng}  = m_X – m_Z = 1.22 – 0,2.23 = 17,4 gam