<p font-family:times="" new="" roman,serif;="" font-size:14.0pt;="" line-height:115%"="">1. CaCO3              --to-->     CaO   +   CO2

2. MgCl2 +2 KOH → Mg(OH)2 +2 KCl
3. Cu(OH)2 +2 HCl → CuCl2 + 2H2O
4. P2O5   +  3H2O    ---->    2 H3PO4
5. 4P +5 O2 to→ 2P2O5
6. 2Al  + 3  H2SO4    ---->     Al2(SO4)3   +  3H2       
7. Zn  + 2 HCl        ---->     ZnCl2  + H2

8. N2 + O2 to→ 2NO

9. 2NO + O2 to→ 2NO2
10.4NO2 + O2 + 2H2O → 4HNO3

1. CaCO3 ----> CaO + CO2

2. MgCl2 + 2KOH → Mg(OH)2 + 2KCl

3. Cu(OH)2 + 2HCl → CuCl2 + 2H2O

4. P2O5 + 3H2O ----> 2H3PO4

5. 4P + 5O2 → 2P2O5

6. 2Al + 3H2SO4 ----> Al2(SO4)3 + 3H2

7. Zn + 2HCl ----> ZnCl2 + H2

8. N2 + O2 → 2NO

9. 2NO + O2 → 2NO2

10.4NO2 + O2 + 2H2O → 4HNO3

\(CaCO_3\underrightarrow{t^o}CaO+CO_2\uparrow\\ MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+KCl\\ Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ 4P+5O_2\underrightarrow{t^o}2P_2O_5\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\\ N_2+O_2\underrightarrow{t^o}2NO\uparrow\\ 2NO+O_2\underrightarrow{t^o}2NO_2\uparrow\\ 4NO_2+O_2+2H_2O\rightarrow4HNO_3\)

1) \(MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+2KCl\)

2) \(Cu\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

3) \(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)

4) FeO + 2HCl --> FeCl₂ + H₂O

\(MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2+2KCl\)

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

1) MgCl₂ + 2KOH → Mg(OH)₂ + 2KCl

2) Cu(OH)₂ + 2HCl → CuCl₂ + 2H₂O

1) MgCl2 + 2KOH --> Mg(OH)2 + 2KCl

2) Cu(OH)2 + 2HCl --> CuCl2 + 2H2O

\(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\\ 2FeI_3\rightarrow2FeI_2+I_2\\ Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

\(a,\left(1\right)Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\left(1:3:2:3\right)\)

\(\left(2\right)Mg+2HCl\rightarrow MgCl_2+H_2\left(1:2:1:1\right)\\ \left(3\right)2FeCl_3+3Ca\left(OH\right)_2\rightarrow2Fe\left(OH\right)_3+3CaCl_2\left(2:3:2:3\right)\)

\(\left(4\right)K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\left(1:2:2:1:1\right)\)

\(b,PTHH:2Cu+O_2\underrightarrow{t^o}2CuO\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Cu}+m_{O_2}=m_{CuO}\\ m_{O_2}=m_{CuO}-m_{Cu}=32-25,6=6,4\left(g\right)\)

giúp bà 20 c bà vô trang t tick 20 lần gieo kèo oke hum, nhìn ngán chớt ><

\(1,MgCl_2+2KOH\to Mg(OH)_2\downarrow+2KCl\\ 2,Cu(OH)_2+2HCl\to CuCl_2+H_2O\\ 3,Cu(OH)_2+H_2SO_4\to CuSO_4+H_2O\\ 4,FeO+2HCl\to FeCl_2+H_2O\\ 5,Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ 6,Cu(NO_3)_2+2NaOH\to Cu(OH)_2\downarrow+2NaNO_3\\ 7,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ 8,N_2+O_2\buildrel{{t^o}}\over\rightleftharpoons 2NO\\ 9,2NO+O_2\to 2NO_2\\ 10,\text {PT sai}\)

Câu 10:

Sửa PT: \(4NO_2+O_2+2H_2O\to 4HNO_3\)