Question

Answer 1

bài 1

C₂H₄+Br₂->C₂H₄Br₂

0,025---0,025----0,025 mol

n _Br2=\(\dfrac{4}{160}\)=0,025 mol

=>V_C2H4=0,025.22,4=0,56l

=>V _CH4 =5,6-0,56=5,04l

 

 

Answer 2

bài 2

C₂H₄+Br₂->C₂H₄Br₂

0,025---------------0,025 mol

n _C2H4Br2=\(\dfrac{4,7}{188}\)=0,025 mol

=>%V_C2H4=\(\dfrac{0,025.22,4}{2,8}\).100=20%

=>%V_CH4=100-20=80%

 

Answer 3

Bài 2 : 

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

Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_2|\)

           1          1             1

         0,025                  0,025

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

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

\(V_{CH4\left(dktc\right)}=2,8-0,56=2,24\left(l\right)\)  

⁰/₀V_CH4 = \(\dfrac{2,24.100}{2,8}=80\)⁰/₀

⁰/₀V_C2H2 = \(\dfrac{0,56.100}{2,8}=20\)⁰/₀

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