PTHH: \(Mg+X_2\rightarrow MgX_2\)

Ta có: \(\dfrac{0,24}{24}=\dfrac{1,84}{24+2\cdot M_X}\) \(\Rightarrow M_X=80\)

Vậy Halogen cần tìm là Brom

       Muối thu được là Magie Bromua

\(n_{K}=\dfrac{3,9}{39}=0,1(mol)\\ PTHH:2K+X_2\to 2KX\\ \Rightarrow n_{K}=n_{KX}=0,1(mol)\\ \Rightarrow M_{KX}=\dfrac{5,8}{0,1}=58(g/mol)\\ \Rightarrow M_{x}=58-39=19(g/mol)\)

Vậy X là Flo (F)

1)

CTHH MX_n

\(n_{X_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

=> \(n_{MX_n}=\dfrac{0,08}{n}\left(mol\right)\)

=> \(n_{AgX}=0,08\left(mol\right)\)

=> \(M_{AgX}=\dfrac{11,48}{0,08}=143,5\left(g/mol\right)\) => M_X = 35,5 (g/mol)

=> X là Cl

2)

\(n_{MCl_n}=\dfrac{0,08}{n}\left(mol\right)\)

\(n_M=\dfrac{0,96}{M_M}\left(mol\right)\)

=> \(\dfrac{0,08}{n}=\dfrac{0,96}{M_M}\)

=> M_M = 12n (g/mol)

Xét n = 1 => Loại

Xét n = 2 => M_M = 24 (g/mol) => M là Mg

Xét n = 3 => Loại

Vậy M là Mg

M^' có hóa trị II

\(n_{O_2}=\dfrac{4,162-0,96-2,242}{32}=0,03\left(mol\right)\)

PTHH: 2Mg + O₂ --t^o--> 2MgO

           0,04-->0,02------>0,04

            2M^' + O₂ --t^o--> 2M^'O

          0,02<-0,01------>0,02

=> M_M' = \(\dfrac{2,242}{0,02}=112\left(g/mol\right)\)

a) \(\left\{{}\begin{matrix}\%n_M=\dfrac{0,04}{0,04+0,02}.100\%=66,67\%\\\%n_{M^{\cdot}}=100\%-66,67\%=33,33\%\end{matrix}\right.\)

b) \(\dfrac{M_M}{M_{M^{\cdot}}}=\dfrac{24}{112}=\dfrac{3}{14}\)

c) \(n_O=0,06\left(mol\right)\)

=> \(n_{H_2O}=0,06\left(mol\right)\)

=> \(n_{H_2SO_4}=0,06\left(mol\right)\)

=> \(C_{M\left(H_2SO_4\right)}=\dfrac{0,06}{0,5}=0,12M\)

lm giúp mình đg cần gấp 

2Al+3X2-->2AlX3

0,3---------------0,3mol

nAl=8,1\27=0,3 mol

=>40,05=0,3.(27+X.3)=>x=35,5 (Cl)

\(n_{X_2}=\dfrac{m}{2.M_X}\left(mol\right)\)

PTHH: Mg + X₂ --> MgX₂

                \(\dfrac{m}{2.M_X}\)-->\(\dfrac{m}{2.M_X}\)

            2Al + 3X₂ --> 2AlX₃

                   \(\dfrac{m}{2.M_X}\)->\(\dfrac{m}{3.M_X}\)

=> \(\left\{{}\begin{matrix}\dfrac{m}{2.M_X}\left(24+2.M_X\right)=19\\\dfrac{m}{3.M_X}\left(27+3.M_X\right)=17,8\end{matrix}\right.\)

=> M_X = 35,5 (g/mol)

=> X là Cl

Đáp án B

Mg + X₂ → MgX₂

2Al + 3X₂→  2AlX₃

Bảo toàn nguyên tố X:

2n_MgX2 = 3n_AlX3 => 2.9,5/(24 + 2X) = 3.8,9/(27 + 31X) => X = 35,5 (Cl)