Gọi 2 kim loại đó là A và B, tên chung 2 kim loại là X. (M(A) < M(X) <M(B) )

PTHH: X+ HCl -> XCl + 1/2 H2

nH2=0,6(mol) -> nX=1,2(mol)

=> M(X)=13,2/1,2=11(g/mol)

=>A là Liti , B là Natri

PTHH: Li + HCl -> LiCl + 1/2 H2

a________a_____a_____0,5a(mol)

Na + HCl -> NaCl + 1/2 H2

b_____b___b______0,5b(mol)

Ta có hpt: \(\left\{{}\begin{matrix}7a+23b=13,2\\0.5a+0,5b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,9\\y=0,3\end{matrix}\right.\)

=> m=m(muối)= mLiCl+ mNaCl= 42,5 . 0,9+ 58,5. 0,3=55,8(g)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ KL:M\\ M+2HCl\rightarrow MCl_2+H_2\\ n_{MCl_2}=n_M=n_{H_2}=0,05\left(mol\right)\\ M_{MCl_2}=\dfrac{4,75}{0,05}=95\left(\dfrac{g}{mol}\right)\\ M\text{à}:M_{MCl_2}=M_M+71\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_M+71=95\\ \Leftrightarrow M_M=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow M\left(II\right):Magie\left(Mg=24\right)\\ a=24.0,05=1,2\left(g\right)\)

Bạn xem lại đề gõ thật chuẩn lại nhé!

Bảo toàn khối lượng: \(m_S=7,8-4,6=3,2\left(g\right)\)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

PTHH: 2X + nS --t^o--> X₂S_n

            \(\dfrac{0,2}{n}\leftarrow0,1\)

\(M_X=\dfrac{4,6}{\dfrac{0,2}{n}}=23n\left(\dfrac{g}{mol}\right)\)

Xét n = 1 thoả mãn => M_X = 23 (g/mol)

=> X là Na

Bài 1:

\(n_M=\dfrac{16}{M_M}\left(mol\right)\)

PTHH: 2M + O₂ --t^o--> 2MO

         \(\dfrac{16}{M_M}\)---------->\(\dfrac{16}{M_M}\)

=> \(\dfrac{16}{M_M}\left(M_M+16\right)=20\)

=> M_M = 64 (g/mol)

=> M là Cu

Bài 2:

\(n_R=\dfrac{16,2}{M_R}\left(mol\right)\)

PTHH: 2R + 3Cl₂ --t^o--> 2RCl₃

          \(\dfrac{16,2}{M_R}\)------------>\(\dfrac{16,2}{M_R}\)

=> \(\dfrac{16,2}{M_R}\left(M_R+106,5\right)=80,1\)

=> M_R = 27 (g/mol)

=> R là Al

 1 
ADDDLBTKL ta có
\(m_{O_2}=m_{MO}-m_M\\ m_{O_2}=20-16=4g\\ n_{O_2}=\dfrac{4}{32}=0,125\left(mol\right)\\ pthh:2M+O_2\underrightarrow{t^o}2MO\) 
            0,25   0,125 
\(M_M=\dfrac{16}{0,25}=64\left(\dfrac{g}{mol}\right)\) 
=> M là Cu 
2 
ADĐLBTKL ta có 
\(m_{Cl_2}=m_{RCl_3}-m_R\\ m_{Cl_2}=80,1-16,2=63,9g\\ n_{Cl_2}=\dfrac{63,9}{71}=0,9\left(mol\right)\\ pthh:2R+3Cl_2\underrightarrow{t^o}2RCl_3\) 
            0,6   0,9 
\(M_R=\dfrac{16,2}{0,6}=27\left(\dfrac{g}{mol}\right)\) 
=> R là Al

\(1 ) 2M+O_2\rightarrow 2MO n_M=n_{MO}\Leftrightarrow \dfrac{16}{M_M}=\dfrac{20}{m_M+16} \Rightarrow m_m = 64(g/mol) \rightarrow M : Cu \)

\(2) 2R+3Cl_2\rightarrow 2RCl_3 n_R=nn_{RCl_3}\Leftrightarrow \dfrac{16,2}{M_R}=\dfrac{80,1}{M_R+35,5.3}\Rightarrow M_R = 27(g/mol)\rightarrow R:Al \)

Đáp án đúng : A

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

\(M+H_2O\rightarrow MOH+\dfrac{1}{2}H_2\\ m_{H_2}=2,3-2,2=0,1\left(g\right)\Rightarrow n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ n_M=2.0,05=0,1\left(mol\right)\\ M_M=\dfrac{2,3}{0,1}=23\left(\dfrac{g}{mol}\right)\\ \Rightarrow M\left(I\right):Natri\left(Na=23\right)\)