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

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)

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

B. dd HCl

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

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

- Kim loại :

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Cu+2FeCl_3\rightarrow CuCl_2+2FeCl_2\)

- Oxit :

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

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

- Hidroxit :

\(Zn\left(OH\right)_2+2HCl\rightarrow ZnCl_2+2H_2O\)

\(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)

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

\(Zn+Cl_2\xrightarrow[]{t^o}ZnCl_2\)

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

\(Cu+Cl_2\xrightarrow[]{t^o}CuCl_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

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

\(Zn\left(OH\right)_2+2HCl\rightarrow ZnCl_2+2H_2O\)

\(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)

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

Câu 2

Ta có

a)n O2=06,72/22,4=0,3(mol)

b) Số phân tử oxi=0,3.6,10\(^{23}=1,8.10^{23}\left(pt\right)\)

c)m O2=0,3.32=9,6(g)

Câu 3

n H2=\(\frac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\)

n SO2=\(\frac{6,4}{64}=0,1\left(mol\right)\)

a) VSO2=(1,5+2,5+0,2+0,1).22,4=96,32(l)

b) m O2=1,5.32=48(g)

m N2=2,5.28=70(g)

m H2=0,2.2=0,4(g)

m SO2=6,4(bài cho)

Tổng khối lượng hỗn hợp là

48+70+0,4+6,4=124,8(g)