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222-9+11+12:2*14+14 = ? ( )

Hóa học Lớp 8: 1. K + O2 -> K2O 2. Al + Cl2 -> AlCl3 3. K + H2O –> KOH + H2 4. Mg +HCl -> MgCl2 + H2 5. Z

Hóa học Lớp 8: 1. K + O2 -> K2O
2. Al + Cl2 -> AlCl3
3. K + H2O –> KOH + H2
4. Mg +HCl -> MgCl2 + H2
5. Zn + HCl -> ZnCl2 + H2
6. Fe2O3 + H2 -> Fe + H2O
7. KMnO4 -> K2MnO4 + MnO2 + O2
8. Na2O + H2O -> NaOH
9. Fe2O3 + CO -> Fe + CO2
10. Al(OH)3 + H2SO4 — > Al2(SO4)3 + H2O
11. Fe3O4 + H2 — > Fe + H2O
12. Fe2O3 + H2 -> Fe + H2O
13. Ca(OH)2 + H3PO4 -> Ca3(PO4)2 + H2O
14. Al + H2SO4 -> Al2(SO4)3 + H2
15. NaOH + FeCl3 -> NaCl + Fe(OH)3, giúp em giải bài hóa này ạ, em cảm ơn thầy cô và các bạn nhiều.

Comments ( 2 )

  1. 1. $4K + O_{2} \rightarrow 2K_{2}O$
    2. $Al + 3Cl_{2} \rightarrow 2AlCl_{3}$
    3. $2K + 2H_{2}O \rightarrow 2KOH + H_{2}$
    4. $Mg + HCl \rightarrow MgCl_{2} + H_{2}$
    5. $Zn + HCl \rightarrow ZnCl_{2} + H_{2}$
    6. $Fe_{2}O_{3} + H_{2} \rightarrow Fe + H_{2}O$
    7. $2KMnO4 \rightarrow K_{2}MnO_{4} + MnO_{2} + O_{2}$
    8. $Na_{2}O + H_{2}O \rightarrow 2NaOH$
    9. $Fe_{2}O_{3} + CO \rightarrow Fe + CO_{2}$
    10. $2Al(OH)_{3} + 3H_{2}SO_{4} \rightarrow Al_{2}(SO_{4})_{3} + 6H_{2}O$
    11. $Fe3O4 + H2 \rightarrow Fe + 4H2O$
    12. $Fe2O3 + H2 -> Fe + H2O$
    13. $Ca(OH)2 + H3PO4 -> Ca3(PO4)2 + H2O$
    14. $Al + H2SO4 -> Al2(SO4)3 + H2$
    15. $NaOH + FeCl3 -> NaCl + Fe(OH)3$

  2. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    1. 4K + $O_{2}$ ⇒2$K_{2}O$
    2. 2Al + 3$Cl_{2}$ ⇒ 2$AlCl_{3}$
    3. K + $H_{2}O$ ⇒ KOH + $H_{2}$
    4. Mg +2HCl ⇒ MgCl2 + $H_{2}$
    5. Zn + 2HCl ⇒ $ZnCl_{2}$ + $H_{2}$
    6. $Fe_{2}O_{3}$ + 3$H_{2}$ ⇒ 2Fe + 3$H_{2}O$
    7. 2$KMnO_{4}$ ⇒ $K_{2}MnO_{4}$+ $MnO_{2}$ + $O_{2}$
    8. $NaO_{2}$ + $H_{2}O$⇒ 2NaOH
    9. $Fe_{2}O_{3}$ + 3CO ⇒ Fe + 3$CO_{2}$
    10. $Al(OH)_{3}$ + 3$H_{2}SO_{4}$⇒ $Al_{2}(SO_{4})_3$ +6$H_{2}O$
    11. $Fe_{3}O_{4}$+ $H_{2}$⇒ Fe + 4$H_{2}O$
    12. $Fe_{2}O_{3}$+ $H_{2}$ ⇒ Fe + $H_{2}O$
    13. 3Ca(OH)2 + 2$H_{3}PO_{4}$ ⇒ $Ca_{3}(SO_{4})_2$ + 6$H_{2}O$
    14. 2Al + 3$H_{2}SO_{4}$ ⇒ $Al_{2}(SO_{4})_3$ + 3$H_{2}$
    15. 3NaOH + FeCl3 ⇒ 3NaCl + $Fe(OH)_{3}$

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222-9+11+12:2*14+14 = ? ( )