Bài 1:

\(a)Mg+2HCl\rightarrow MgCl_2+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ b)Zn+2HCl\rightarrow ZnCl_2+H_2\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Bài 2:

\(a)Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b)n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ n_{Fe}=n_{H_2SO_4}=n_{H_2}=0,15mol\\ m_{Fe}=0,15.56=8,4g\\ c)C_{M_{H_2SO_4}}=\dfrac{0,15}{0,05}=3M\)

khó quá

PTHH: Na₂CO₃ + Ca(OH)₂ → CaCO₃↓ + 2NaOH

Bài 1:

a. Zn + 2HCl -> ZnCl₂ + H₂

b. CuO + 2HCl -> CuCl₂ + H₂O

c. Ba(OH)₂ + 2HCl - > BaCl₂ + 2H₂O

d. Fe(OH)₃ + 3HCl -> FeCl₃ + 3H₂O

B1:

\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,CuO+2HCl\rightarrow CuCl_2+H_2O\\ c,Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\\d, Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

B2:

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{HCl}=0,1.3=0,3\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{2}\Rightarrow HCldư\\ b,n_{H_2}=n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ n_{HCl\left(dư\right)}=0,3-0,1.2=0,1\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,C_{MddMgCl_2}=\dfrac{0,1}{0,1}=1\left(M\right)\\ C_{MddHCl\left(dư\right)}=\dfrac{0,1}{0,1}=1\left(M\right)\)

\(a.CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ CO_2+H_2O⇌H_2CO_3\\ Na_2O+H_2O\rightarrow2NaOH\\ SO_3+H_2O\rightarrow H_2SO_4\\ b.CaO+2HCl\rightarrow CaCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Na_2O+2HCl\rightarrow2NaCl+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ c.CO_2+NaOH\rightarrow NaHCO_3\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ SO_3+NaOH\rightarrow NaHSO_4\\ SO_3+2NaOH\rightarrow Na_2SO_4+H_2O\)

$PTHH:Zn+2HCl\to ZnCl_2+H_2\uparrow$

a)

$SO_3 + H_2O \to H_2SO_4$
$K_2O + H_2O \to 2KOH$
$CO_2 + H_2O \rightleftharpoons H_2CO_3$

b)

$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$
$K_2O + 2HCl \to 2KCl + H_2O$

c)

$2NaOH + SO_3 \to Na_2SO_4 + H_2O$
$2NaOH + CO_2 \to Na_2CO_3 + H_2O$

a) Tạo ra chất rắn không tan.

b) Canxi hiđroxit + Khí cacbon đioxit ­---- > Canxi cacbonat + Nước.

a)

$MgO$ : Magnesium oxide

$Mg(OH)_2$ : Magnesium hydroxide

b)

$2Mg + O_2 \xrightarrow{t^o} 2MgO$

$Mg + Cl_2 \xrightarrow{t^o} MgCl_2$

$Mg + 2HCl \to MgCl_2 + H_2$

sắt cháy mãnh liệt và bắn ra vài hạt vụn 
3Fe + 2O₂ -t^o-> Fe₃O₄ 
đồng chuyển từ màu đen sang màu đỏ 
CuO + H₂  -t^o-> Cu + H₂O 
kẽm sủi bọt và giải phóng khí hidro 
Zn + 2HCl -> ZnCl₂ + H₂