a) 

C₂H₄ + Br₂ --> C₂H₄Br₂

C₂H₂ + 2Br₂ --> C₂H₂Br₄

b) Gọi số mol C₂H₄, C₂H₂ là a, b (mol)

=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (1)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

                a---->a

             C₂H₂ + 2Br₂ --> C₂H₂Br₄

                b---->2b

=> a + 2b = \(\dfrac{112}{160}=0,7\) (2)

(1)(2) => a = 0,3 (mol); b = 0,2 (mol)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,5}.100\%=40\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,3.28}{0,3.28+0,2.26}.100\%=61,765\%\\\%m_{C_2H_2}=\dfrac{0,2.26}{0,3.28+0,2.26}.100\%=38,235\%\end{matrix}\right.\)

a)

$C_2H_2 + 2Br_2 \to C_2H_2Br_4$

b) $n_{Br_2} = \dfrac{40.20\%}{160} = 0,05(mol)$

Theo PTHH : $n_{C_2H_2Br_4} = n_{C_2H_2} = \dfrac{1}{2}n_{Br_2} = 0,025(mol)$

$m_{C_2H_2Br_4} = 0,025.346 = 8,65(gam)$

c) $\%V_{C_2H_2} = \dfrac{0,025}{0,5}.100\% = 5\%$

$\%V_{CH_4} = 100\% - 5\% = 95\%$

n Br2=\(\dfrac{32}{160}\)=0,2 mol

C2H2+2Br2->C2H2Br4

0,1------0,2 mol

=>%VC2H2=\(\dfrac{0,1.22,4}{5,6}\).100=40%

=>%VCH4=100-40=60%

=>n CH4=\(\dfrac{5,6-0,1.22,4}{22,4}\)=0,15 mol

CH4+2O2-to>CO2+2H2O

0,15----0,3

C2H2+\(\dfrac{5}{2}\)O2-to>2CO2+H2O

0,1-----0,25 mol

=>VO2=(0,3+0,25).22,4=12,32l

a, Cho hỗn hợp khí CH₄ và C₂H₄ qua dung dịch Br₂ dư chỉ có C₂H₄ tham gia phản ứng. PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b, \(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{64}{160}=0,4(mol)\)

Theo PTHH: \(n_{C_2H_4}=n_{Br_2}=0,4(mol)\)

\(V_{C_2H_4}=n_{C_2H_4}.22,4=0,4.22,4=8,96(l)\)

Phần trăm của khí C₂H₄ trong hỗn hợp ban đầu là: \(\%V_{C_2H_4}=\dfrac{V_{C_2H_4}}{V_hh}.100\%=\dfrac{8,96}{11,2}.100\%=80\%\)

Phần trăm của khí CH₄ trong hỗn hợp ban đầu là: 

\(\%V_{CH_4}=100\%-\%V_{C_2H_4}=100\%-80\%=20\%\)

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

\(n_{Br_2}=\dfrac{2,4}{160}=0,015mol\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

0,0075  0,015                       ( mol )

\(V_{C_2H_2}=0,0075.22,4=0,168l\)

\(V_{CH_4}=3,36-0,168=3,192l\)

\(\%V_{C_2H_2}=\dfrac{0,168}{3,36}.100=5\%\)

\(\%V_{CH_4}=100\%-5\%=95\%\)

giúp vs nha cần lắm!!!

a, \(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)

PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

_____0,15____0,3 (mol)

\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,15.22,4}{11,2}.100\%=30\%\)

\(\Rightarrow\%V_{CH_4}=100-30=70\%\)

b, - Khí thoát ra ngoài là CH₄.

\(V_{CH_4}=11,2.70\%=7,84\left(l\right)\)

Sửa đề : 11.2 (l) 

\(n_{hh}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(m_{Br_2}=320\cdot\dfrac{15}{100}=48\left(g\right)\)

\(n_{Br_2}=\dfrac{48}{160}=0.3\left(mol\right)\)

\(\)\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

\(0.15..........0.3\)

\(V_{CH_4}=0.5-0.15=0.35\left(mol\right)\)

\(\%C_2H_2=\dfrac{0.15}{0.5}\cdot100\%=30\%\)

\(\%CH_4=70\%\)