a_ \(2HCl-^{đpdd}\rightarrow Cl_2+H_2\)

\(\dfrac{3}{2}Cl_2+Fe-^{t^o}\rightarrow FeCl_3\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(NaCl_{\left(r,k\right)}+H_2SO_{4\left(đ,n\right)}-^{t^o}\rightarrow NaHSO_4+HCl\)

\(2HCl+CuO\rightarrow CuCl_2+H_2O\)

\(CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\)

a_ 2HCl−đpdd→Cl2+H22HCl−đpdd→Cl2+H2

\(2KMnO_4+16HCl_đ-^{t^o}\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(Cl_2+H_2-^{as}\rightarrow2HCl\)

\(6HCl+Fe_2O_3\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\)

\(2AgCl-^{t^o}\rightarrow2Ag+Cl_2\)

\(Cl_2+2NaBr\rightarrow2NaCl+Br_2\)

\(Br_2+2NaI\rightarrow2NaBr+I_2\)

\(I_2+Zn-^{t^o}\rightarrow ZnI_2\)

\(ZnI_2+2NaOH_{\left(vđ\right)}\rightarrow Zn\left(OH\right)_2+2NaI\)

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{13}{65}=0,2\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\\ b,n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ Theo.PTHH:\) 

0,2____0,4______0,2____0,2

\(\rightarrow V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

a) 

2KMnO4 + 16HCl => 2KCl + 2MnCl2 + 5Cl2 + 8H2O 

6KOH + 3Cl2 -to-> 5KCl + KClO3 + 3H2O 

KClO3 + 6HCl  => KCl + 3Cl2  + 3H2O 

2KCl + 2H2O -dpcmn-> 2KOH + H2  + Cl2 

b) 

BaCl2 -dpnc-> Ba + Cl2 

Cl2 + H2 -to-> 2HCl 

Fe + 2HCl => FeCl2 + H2 

FeCl2 + 1/2Cl2 => FeCl3 

2FeCl3 + 3Ba(OH)2 => 2Fe(OH)3 + 3BaCl2

BaCl2 + H2SO4 => BaSO4 + 2HCl 

Chúc bạn học tốt , có thắc mắc thì cứ hỏi nhé !!

a)

$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$

b)

n HCl = 0,4.1,5 = 0,6(mol)

n Al2O3 = 1/6 n HCl = 0,1(mol) => m = 0,1.102 = 10,2(gam)

n AlCl3 = 1/3 n HCl = 0,2(mol) => CM AlCl3 = 0,2/0,4 = 0,5M

nHCl = 0.4*1.5 = 0.6 (mol) 

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

nAlCl3 = 0.6*2/6 = 0.2 (mol) 

mAlCl3 = 0.2*133.5 = 26.7 (g) 

CM AlCl3 = 0.2/0.4 = 0.5 (M) 

Sửa đề: 7,2% → 7,3%

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

\(m_{HCl}=100.7,3\%=7,3\left(g\right)\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,2}{2}\), ta được HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{H_2}=n_{Fe}=0,05\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{FeCl_2}=0,05.127=6,35\left(g\right)\)

\(m_{H_2}=0,05.2=0,1\left(g\right)\)

\(n_{HCl\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)