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

Hóa học Lớp 8: 1. Fe2O3 + CO ==–> Fe + CO2 2. AgNO3 + Al ==–>

Hóa học Lớp 8: 1. Fe2O3 + CO ——–> Fe + CO2
2. AgNO3 + Al ——–> Al(NO)3 + Ag
3. HCl + CaCO3 ——–> CaCl2 + H2O
4. C4H10 + O2 ——–> CO2 + H2O
5. NaOH + Fe2(SO4)3 ——–> Fe(OH)3 + Na2SO4
6. FeS2 + O2 ——-> Fe2O3 + SO2
7. KOH + Al2(SO4)3 —–> K2SO4 + Al(OH)3
8. CH4 + O2 + H2O ——> CO2 + H2
9. Al + Fe3O4 ——> Al2O3 + Fe
10. FexOy + CO ——> FeO + CO2
1. Fe2O3 + CO —–> FexOy + CO2
2. KMnO4 ——> K2MnO4 + MnO2 + O2
3. Al + FexOy ——-> Fe + Al2O3
4. Fe + O2 —— > FexOy
5. Na2O + H2O ——-> NaOH
6. Zn + HCl ——> ZnCl2 + H2
7. CuO + H2 ——>Cu + H2O
8. FeS + O2 ——> Fe2O3 + SO2
12. KOH + Al2(SO4)3 ——> K2SO4 +Al(OH)3
14. FeS2 + O2 ——-> Fe2O3 + SO2
14. CaO +H3PO4 ——–> Ca3(PO4)2 + H2O
15. Fe3O4 + HCl ———-> FeCl3 + FeCl2 + H2O
17. Fe2O3 + CO ———–> Fe + CO2
19. CaCO3 + HCl ——–> CaCl2 + H2O + CO2
20. NaOH + Fe2(SO4)3 —–> Fe(OH)3 + Na2SO4
21. FeS2 + O2 ——-> Fe2O3 + SO2
23. Al + Fe3O4 —> Fe + Al2O3
24. FexOy + CO —-> FeO + CO2, giúp em giải bài hóa này ạ, em cảm ơn thầy cô và các bạn nhiều.

Comments ( 1 )

  1. Giải đáp:
    Bạn tham khảo lời giải ở dưới nhé!!!
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    1,F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
    2,Al + 3AgN{O_3} \to Al{(N{O_3})_3} + 3Ag\\
    3,CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
    4,2{C_4}{H_{10}} + 13{O_2} \to 8C{O_2} + 10{H_2}O\\
    5,F{e_2}{(S{O_4})_3} + 6NaOH \to 2Fe{(OH)_3} + 3N{a_2}S{O_4}\\
    6,4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}\\
    7,A{l_2}{(S{O_4})_3} + 6KOH \to 2Al{(OH)_3} + 3{K_2}S{O_4}\\
    8,2C{H_4} + {O_2} + 2{H_2}O \to 2C{O_2} + 4{H_2}\\
    9,8Al + 3F{e_3}{O_4} \to 4A{l_2}{O_3} + 9Fe\\
    10,F{e_x}{O_y} + (x – y)CO \to xFeO + (x – y)C{O_2}\\
    11,xF{e_2}{O_3} + (3x – 2y)CO \to 2F{e_x}{O_y} + (3x – 2y)C{O_2}\\
    12,2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
    13,2yAl + 3F{e_x}{O_y} \to 3xFe + yA{l_2}{O_3}\\
    14,2xFe + y{O_2} \to 2F{e_x}{O_y}\\
    15,N{a_2}O + {H_2}O \to 2NaOH\\
    16,Zn + 2HCl \to ZnC{l_2} + {H_2}\\
    17,CuO + {H_2} \to Cu + {H_2}O\\
    18,4FeS + 7{O_2} \to 2F{e_2}{O_3} + 4S{O_2}\\
    19,A{l_2}{(S{O_4})_3} + 6KOH \to 2Al{(OH)_3} + 3{K_2}S{O_4}\\
    20,4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}\\
    21,3CaO + 2{H_3}P{O_4} \to C{a_3}{(P{O_4})_2} + 3{H_2}O\\
    22,F{e_3}{O_4} + 8HCl \to 2FeC{l_3} + FeC{l_2} + 4{H_2}O\\
    23,F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
    24,CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
    25,F{e_2}{(S{O_4})_3} + 6NaOH \to 2Fe{(OH)_3} + 3N{a_2}S{O_4}\\
    26,4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}\\
    27,8Al + 3F{e_3}{O_4} \to 4A{l_2}{O_3} + 9Fe\\
    28,F{e_x}{O_y} + (x – y)CO \to xFeO + (x – y)C{O_2}
    \end{array}\)

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

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