Question

giúp mk vs ạ

Answer 1

a) \(n_{C_2Ag_2}=\dfrac{24}{240}=0,1\left(mol\right)\)

=> n_C2H2 = 0,1 (mol)

=> V_C2H4 = 3,36 - 0,1.22,4 = 1,12 (l)

\(\%V_{C_2H_4}=\dfrac{1,12}{3,36}.100\%=33,33\%\)

b) 
\(n_{C_2H_4}=\dfrac{6,72.33,33\%}{22,4}=0,1\left(mol\right)\)

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

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

               0,1->0,1

            C₂H₂ + 2Br₂ --> C₂H₂Br₄

              0,2--->0,4

=> m_Br2 = (0,1 + 0,4).160 = 80(g)