Question

Answer 1

Gọi halogen cần tìm là X

\(n_{MgX_2}=\dfrac{19}{24+2.M_X}\left(mol\right)\)

\(n_{AlX_3}=\dfrac{17,8}{27+3.M_X}\left(mol\right)\)

PTHH: Mg + X₂ ---> MgX₂

            \(\dfrac{19}{24+2.M_X}\)<-\(\dfrac{19}{24+2.M_X}\)

             2Al + 3X₂ --> 2AlX₃

                 \(\dfrac{17,8}{18+2.M_X}\)<-\(\dfrac{17,8}{27+3.M_X}\)

=> \(\dfrac{19}{24+2.M_X}=\dfrac{17,8}{18+2.M_X}\)

=> M_X = 35,5 (g/mol)

=> X là Cl (Clo)

\(n_{Cl_2}=\dfrac{19}{24+2.35,5}=0,2\left(mol\right)\)

=> m_Cl2 = 0,2.71 = 14,2 (g)