Bài 14 : 

Vì metan không tác dụng với Brom nên : 

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

a) Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_{2|}\)

              1          1                1

            0,025                    0,025

b) \(n_{C2H4}=\dfrac{0,025.1}{1}=0,025\left(mol\right)\)

\(V_{C2H4\left(dktc\right)}=0,025.22,4=0,56\left(l\right)\)

\(V_{CH4\left(dktc\right)}=1,4-0,56=0,84\left(l\right)\)

⁰/₀V_CH4 = \(\dfrac{0,84.100}{1,4}=60\)⁰/₀

⁰/₀V_C2H4 = \(\dfrac{0,56.100}{1,4}=40\)⁰/₀

 Chúc bạn học tốt

Thanks ạ

nC2H4Br2 = \(\dfrac{4,7}{188}=0,025\left(mol\right)\)

C2H4 + Br2 -> C2H4Br2

0,025 <-----------0,025

=>VC2H4 = 0,025 . 22,4=0,56(l)

=> VCH4 = 2,8 - 0,56 =2,24 (l)

%VCH4 = \(\dfrac{2,24.100}{2,8}=80\%\)

%VC2H4 = 100 % -80% = 20%

n_C2H4Br2 = 0,025 mol

C₂H₄ + Br₂ → C₂H₄Br₂

⇒ V_C2H4 = 0,025.22,4 = 0,56 (l)

⇒ %_C2H4 = \(\dfrac{0,56.100\%}{2,8}\) = 20%

⇒ %_CH4 = 100% - 20% = 80%

1. \(n_{Br_2}=0,4.0,5=0,2\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)

2. \(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,05\left(mol\right)\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,05.22,4}{1,4}.100\%=80\%\)

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

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{C_2H_4Br_2} = \dfrac{4,7}{188} = 0,025(mol)\\ V_{C_2H_4} = 0,025.22,4 = 0,56(lít)\)

Ta có:

nhh = 0,125(mol)

=> nC2H4Br2 = 4,7/188 = 0,025(mol)

C2H4 + Br2 => C2H4Br2

0,025_______0,025__________

=> nCH4 = 0,125-0,025 = 0,1(mol)

=> %VCH4 = 0,1.100/0,125 = 80%

a) \(n_{C_2H_4Br_2}=\dfrac{4,7}{188}=0,025\left(mol\right)\)

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

           0,025<------------0,025

b) \(\left\{{}\begin{matrix}V_{C_2H_4}=0,025.22,4=0,56\left(l\right)\\V_{CH_4}=5,6-0,56=5,04\left(l\right)\end{matrix}\right.\)

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

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

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

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

a) \(C_2H_4 + Br_2 \to C_2H_4Br_2\)

b)

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

Cho 1,72 lít hỗn hợp khí gồm metan và etilen (đktc) lội qua dd brom dư lượng brom tham gia phản ứng là 11,2 gam

a. Viết PTHH?

b. Tính thành phần phần trăm về thể tích mỗi khí trong hỗn hợp ?

Ta có:

nhh = 0,125(mol)

=> nC2H4Br2 = 4,7/188 = 0,025(mol)

C2H4 + Br2 => C2H4Br2

0,025_______0,025__________

=> nCH4 = 0,125-0,025 = 0,1(mol)

=> %VCH4 = 0,1.100/0,125 = 80%

=> %VC2H4 = 100% - 80% = 20%