Bài 1 : 

$a) Fe + 2HCl \to FeCl_2 + H_2$
$b) Cu(OH)_2 + 2HCl \to CuCl_2 + 2H_2O$

$c)  Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$
$d) CuCl_2 + Fe \to FeCl_2 + Cu$

Bài 2 : 

$CaO  + H_2O \to Ca(OH)_2$
$Na_2CO_3 + Ca(OH)_2 \to CaCO_3 + 2NaOH$

Bài 1: \(a,Fe+2HCl\rightarrow FeCl_2+H_2\)

\(b,Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\)

\(c,Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

\(d,Cu\left(OH\right)_2\underrightarrow{t^0}CuO+H_2O\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)

Tham khảo

Bài 1

a) Fe+2HCl\(\rightarrow\)FeCl₂+H₂

b) Cu(OH)₂+2HCl\(\rightarrow\)CuCl₂+2H₂O

c) Na₂CO₃+2HCl\(\rightarrow\)NaCl+CO₂+H₂O

d) Fe+CuCl₂\(\rightarrow\)FeCl₂+Cu

Bài 2:

CaO+H₂O\(\rightarrow\)Ca(OH)₂

Ca(OH)₂+Na₂CO₃\(\rightarrow\)CaCO₃+2NaOH

Bài 3

a) Na₂O+H₂O\(\rightarrow\)2NaOH

b) BaO+H₂O\(\rightarrow\)Ba(OH)₂

c) Ba(OH)₂+CuSO₄\(\rightarrow\)BaSO₄+Cu(OH)₂

d) CuSO₄+2NaOH\(\rightarrow\)Cu(OH)₂+Na₂SO₄

e) FeCl₂+2NaOH\(\rightarrow\)Fe(OH)₂+2NaCl

a. Fe + 2HCl = FeCl2 + H2

2NaCl + 2H2O -- đpdd có mn --> 2NaOH + H2 + Cl2

3Cl2 + 2Fe --t^o --> 2FeCl3

2FeCl3 + Cu --> 2FeCl2 + CuCl2

b. Cu(OH)2 + 2HCl = CuCl2 + 2H2O

c. Na2CO3 + 2HCl = NaCl + CO2 + H2O

d. CuCl2 -- đpdd --> Cu + Cl2

B. dd HCl

Fe phản ứng hoàn toàn với HCl, Cu không phản ứng học lấy.

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Bài 2:

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(Ca\left(OH\right)_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaOH\)

Bài 3

a) 

\(Na_2O+H_2O\rightarrow2NaOH\)

b)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

c) 

\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

\(Na_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2NaOH\)

d) 

\(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

e)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

a) $Zn + 2HCl \to ZnCl_2 + H_2$

b) Theo PTHH : $n_{H_2} = n_{Zn} = \dfrac{16,25}{65} = 0,25(mol)$
$\Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)$

c) $n_{HCl} = 2n_{Zn} = 0,5(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,5}{0,5} = 1M$