\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ b.n_{H_2}=n_{Fe}=\dfrac{10}{56}=\dfrac{5}{28}\left(mol\right)\\ \Rightarrow V_{H_2}=\dfrac{5}{28}.22,4=4\left(l\right)\)

\(a.n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2        0,4          0,2           0,2

\(V_{H_2}=0,2.24,79=4,958l\\ b.m_{HCl}=0,4.36,5=14,6g\\ c.m_{FeCl_2}=0,2.127=25,4g\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

PTHH: Fe + 2 HCl -> FeCl2 + H2

nFe= 0,05(mol) -> nHCl=0,1(mol)  ; nFeCl2=0,05(mol)

mHCl=0,1.36,5=3,65(g)

mFeCl2=127.0,05=6,35(g)

\(n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ 0,03....0,06.....0,03.......0,03\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,03.22,4=0,672\left(l\right)\\ b,m_{HCl}=0,06.36,5=2,19\left(g\right)\\ c,m_{FeCl_2}=127.0,03=3,81\left(g\right)\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$

$n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$

b) $n_{HCl} = 2n_{H_2} = 0,3(mol)$
$\Rightarrow m_{HCl} = 0,3.36,5 = 10,95(gam)$

c) 

Cách 1 : $n_{FeCl_2} = n_{H_2} = 0,15(mol) \Rightarrow m_{FeCl_2} = 0,15.127 = 19,05(gam)$

Cách 2 : Bảo toàn khối lượng, $m_{FeCl_2} = 8,4 + 10,95 - 0,15.2 = 19,05(gam)$

Cíuuuu mn oiii

Fe + 3/2Cl2 -to-> FeCl3

Fe + 2HCl --> FeCl2 + H2

3Fe + 2O2 -to-> Fe3O4

Fe + H2SO4 --> FeSO4 + H2

2Fe + 6H2SO4(đ) -to-> Fe2(SO4)3 + 3SO2 + 6H2O

Al + 3/2Cl2 -to-> AlCl3

2Al + 6HCl --> 2AlCl3 + 3H2

2Al + 3O2 -to-> 2Al2O3

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

2Al + 6H2SO4 (đ) -to-> Al2(SO4)3 + 3SO2 + 6H2O

Zn + Cl2 -to-> ZnCl2

Zn + 2HCl --> ZnCl2 + H2

Zn + 1/2O2 -to-> ZnO

Zn + H2SO4 --> ZnSO4 + H2

Zn + 2H2SO4(đ) -to-> ZnSO4 + SO2 + 2H2O

PTHH: 2Fe +3 Cl₂ \(\underrightarrow{t^o}\) 2FeCl₃
Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂\(\uparrow\)
Fe + H₂SO_{4 loãng} \(\rightarrow\) FeSO₄ + H₂\(\uparrow\)
2Fe + 6H₂SO_{4 đặc} \(\underrightarrow{t^o}\) Fe₂(SO₄)₃ + 6H₂O + 3SO₂\(\uparrow\)
2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂\(\uparrow\)
4Al + 3O₂ \(\underrightarrow{t^o}\) 2Al₂O₃
2Al + 6H₂SO_{4 đặc} \(\underrightarrow{t^o}\) Al₂(SO₄)₃ + 6H₂O + 3SO₂\(\uparrow\)
2Al + 3H₂SO_{4 loãng} \(\rightarrow\) Al₂(SO₄)₃ + 3H₂\(\uparrow\)
Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂\(\uparrow\)
Zn + H₂SO_{4 loãng} \(\rightarrow\) ZnSO₄ + H₂\(\uparrow\)
Zn + 2H₂SO_{4 đặc} \(\rightarrow\) ZnSO₄ + 2H₂O + SO₂\(\uparrow\)

a. KClO₃ + 6HCl → 3Cl₂ + KCl + 3H₂O.

2Fe + 3Cl₂ \(\underrightarrow{t^o}\) 2FeCl₃.

FeCl₃ + 3NaOH \(\rightarrow\) Fe(OH)₃ + 3NaCl.

2Fe(OH)₃ + 3H₂SO₄ \(\rightarrow\) Fe₂(SO₄)₃ + 6H₂O.

b. KClO₃ + 6HCl → 3Cl₂ + KCl + 3H₂O.

2Na + Cl₂ \(\underrightarrow{t^o}\) 2NaCl.

2NaCl + H₂SO₄ (đặc) \(\underrightarrow{250^oC}\) Na₂SO₄ + 2HCl.

2HCl + CuO \(\rightarrow\) CuCl₂ + H₂O.

CuCl₂ + 2AgNO₃ \(\rightarrow\) 2AgCl + Cu(NO₃)₂.

2AgCl + Cu \(\rightarrow\) CuCl₂ + 2Ag.