a) Phương trình hóa học phản ứng 

A + 2Cl ---> ACl₂

1 :  2      :   1         (1)

Ta có \(n_{Cl}=\frac{m}{M}=\frac{2,13}{35,5}=0,06\left(mol\right)\)  (2)

Từ (1) và (2) => n_A = 0,03

=> \(M_A=\frac{m}{n}=\frac{0,72}{0,03}=24\)

=> A là Mg 

=> CTHH của muối B là MgCl₂

b) Ta có n_B = 0,03

=>  \(m_B=n.M=0,03.95=2,85\left(g\right)\)

\(a,PTHH:Zn+Cl_2\rightarrow ZnCl_2\)

\(b,n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\\ Theo.PTHH:n_{Cl_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ a=m_{Cl_2}=n.M=0,4.35,5=14,2\left(g\right)\)

\(b=m_{ZnCl_2}=n.M=0,2.136=27,2\left(g\right)\)

\(c,PTHH:2Al+3Cl_2\rightarrow2AlCl_3\\ Theo.PTHH:n_{Al}=\dfrac{2}{3}.n_{Cl_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ m_{Al}=n.M=\dfrac{2}{15}.27=3,6\left(g\right)\)

Gọi KL cần tìm là M

\(n_{AgNO_3}=\dfrac{170}{170}=1(mol)\\ MCl_2+2AgNO_3\to M(NO_3)_2+2AgCl\downarrow\\ \Rightarrow n_{MCl_2}=\dfrac{1}{2}n_{AgNO_3}=0,5(mol)\\ \Rightarrow M_{MCl_2}=\dfrac{55,5}{0,5}=111(g/mol)\\ \Rightarrow M_M=111-35,5.2=40(g/mol)(Ca)\\ n_{Ca(NO_3)_2}=0,5(mol);n_{AgCl}=1(mol)\\ \Rightarrow m_{Ca(NO_3)_2}=0,5.164=82(g);m_{AgCl}=1.143,5=143,5(g)\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

\(PTHH:R+2HCl\rightarrow RCl_2+H_2\uparrow\\ Mol:0,2\leftarrow0,4\rightarrow0,2\rightarrow0,2\)

=> M_R = \(\dfrac{13}{0,2}=65\left(\dfrac{g}{mol}\right)\)

=> R là Zn

=> \(\left\{{}\begin{matrix}V_{H_2}=0,2.22,4=4,48\left(l\right)\\m_{ZnCl_2}=0,2.136=27,2\left(g\right)\end{matrix}\right.\)

nHCL = 14,6  : 36,5 = 0,4 (MOL) 
pthh :  2R + 2xHCl ---> 2RClx  + xH2 
            0,4x<--0,4 (mol) 
M_R =  13:0,4x = 32,5x(g/mol) 
xét 
x = 1 (KTM ) 
x= 2 (TM ) 
x = 3 (KTM ) 
x =4( KTM ) 
x= 5 (ktm ) 
x=6 (ktm) 
x=7 (ktm ) 
=> R là zn

nHCL = 14,6  : 36,5 = 0,4 (MOL) 
pthh :  2R + 2xHCl ---> 2RClx  + xH2 
            0,4x<--0,4 (mol) 
MR =  13:0,4x = 32,5x(g/mol) 
xét 
x = 1 (KTM ) 
x= 2 (TM ) 
x = 3 (KTM ) 
x =4( KTM ) 
x= 5 (ktm ) 
x=6 (ktm) 
x=7 (ktm ) 
=> R là zn

\(n_{muối}=0.25\cdot3=0.75\left(mol\right)\)

\(M+Cl_2\underrightarrow{t^0}MCl_2\)

\(0.75....0.75...0.75\)

\(M_M=\dfrac{48.75}{0.7}=65\)

\(Mlà:Zn\)

\(V_{Cl_2}=0.75\cdot22.4=16.8\left(l\right)\)

\(a,PTHH:R+2AgNO_3\to R(NO_3)_2+2Ag\\ \Rightarrow n_{R}=n_{R(NO_3)_2}\\ \Rightarrow \dfrac{2,8}{M_R}=\dfrac{9}{M_R+124}\\ \Rightarrow M_R=56(g/mol)\)

Vậy R là sắt (Fe)

\(b,n_{R}=\dfrac{2,8}{56}=0,05(mol)\\ \Rightarrow n_{AgNO_3}=0,1(mol)\\ \Rightarrow m_{dd_{AgNO_3}}=\dfrac{0,1.170}{5\%}=340(g)\\ c,n_{Fe(NO_3)_2}=n_{Fe}=0,05(mol);n_{Ag}=0,1(mol)\\ \Rightarrow C\%_{Fe(NO_3)_2}=\dfrac{0,05.180}{2,8+340-0,1.108}.100\%=2,71\%\)

Câu 1:

\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)

Mg+Br2->MgBr2

0,07--0,07----0,07

n Br2=\(\dfrac{11,2}{160}\)=0,07 mol

=>m Mg=0,07.24=1,68g

=>m MgBr2=0,07.184=12,88g

Câu 2:

\(n_{Al}=\dfrac{14,58}{27}=0,54\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{AlCl_3}=n_{Al}=0,54\left(mol\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,54=0,81\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,81.22,4=18,144\left(l\right)\\ m_{AlCl_3}=0,54.133,5=72,09\left(g\right)\)