2026年突唯中考中考总复习九年级化学全一册通用版河南专版


注:目前有些书本章节名称可能整理的还不是很完善,但都是按照顺序排列的,请同学们按照顺序仔细查找。练习册 2026年突唯中考中考总复习九年级化学全一册通用版河南专版 答案主要是用来给同学们做完题方便对答案用的,请勿直接抄袭。



《2026年突唯中考中考总复习九年级化学全一册通用版河南专版》

一、常见酸($\boldsymbol{HCl}$、$\boldsymbol{{H_{2}SO_{4}}}$)的化学性质(“酸五条”)

1. 请写出下列物质的化学式:A.
NaOH
$\boldsymbol{NaOH}$;B.
Na₂CO₃
$\boldsymbol{{Na_{2}CO_{3}}}$。
2. 请写出图中涉及反应的化学方程式
①_____、_____、_____、_____;
②_____、_____、_____、_____;
③_____、_____;
④_____、_____、_____、_____;
⑤_____;
⑥_____。
答案: 1. A:$\boldsymbol{NaOH}$;B:$\boldsymbol{{Na_{2}CO_{3}}}$。
2. ①$\boldsymbol{{Zn + 2HCl = ZnCl_{2} + H_{2}\uparrow}}$、$\boldsymbol{{Fe + 2HCl = FeCl_{2} + H_{2}\uparrow}}$、$\boldsymbol{{Zn + H_{2}SO_{4} = ZnSO_{4} + H_{2}\uparrow}}$、$\boldsymbol{{Fe + H_{2}SO_{4} = FeSO_{4} + H_{2}\uparrow}}$;
②$\boldsymbol{{HCl + NaOH = NaCl + H_{2}O}}$、$\boldsymbol{{H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O}}$、$\boldsymbol{{2HCl + Ca(OH)_{2} = CaCl_{2} + 2H_{2}O}}$、$\boldsymbol{{H_{2}SO_{4} + Ca(OH)_{2} = CaSO_{4} + 2H_{2}O}}$;
③$\boldsymbol{{2HCl + Na_{2}CO_{3} = 2NaCl + H_{2}O + CO_{2}\uparrow}}$、$\boldsymbol{{H_{2}SO_{4} + Na_{2}CO_{3} = Na_{2}SO_{4} + H_{2}O + CO_{2}\uparrow}}$;
④$\boldsymbol{{6HCl + Fe_{2}O_{3} = 2FeCl_{3} + 3H_{2}O}}$、$\boldsymbol{{Fe_{2}O_{3} + 3H_{2}SO_{4} = Fe_{2}(SO_{4})_{3} + 3H_{2}O}}$、$\boldsymbol{{2HCl + CuO = CuCl_{2} + H_{2}O}}$、$\boldsymbol{{H_{2}SO_{4} + CuO = CuSO_{4} + H_{2}O}}$;
⑤$\boldsymbol{{H_{2}SO_{4} + BaCl_{2} = BaSO_{4}\downarrow + 2HCl}}$;
⑥$\boldsymbol{{HCl + AgNO_{3} = AgCl\downarrow + HNO_{3}}}$。
3. 金属除锈的化学方程式为
$\boldsymbol{{Fe_{2}O_{3} + 6HCl = 2FeCl_{3} + 3H_{2}O}}$(或$\boldsymbol{{Fe_{2}O_{3} + 3H_{2}SO_{4} = Fe_{2}(SO_{4})_{3} + 3H_{2}O}}$),生活中常用含${Mg(OH)_{2}}$的药物以及${NaHCO_{3}}$治疗胃酸过多症的化学方程式分别为
$\boldsymbol{{Mg(OH)_{2} + 2HCl = MgCl_{2} + 2H_{2}O}}$、$\boldsymbol{{NaHCO_{3} + HCl = NaCl + H_{2}O + CO_{2}\uparrow}}$。
答案: 3.Fe₂O₃ + 6HCl = 2FeCl₃ + 3H₂O(合理即可)
Mg(OH)₂ + 2HCl = MgCl₂ + 2H₂O
NaHCO₃ + HCl = NaCl + H₂O + CO₂↑
二、常见碱[$\boldsymbol{NaOH}$、$\boldsymbol{{Ca(OH)_{2}}}$]的化学性质(“碱四条”)

1. 请写出下列物质的化学式:A.
CO₂
$\boldsymbol{{CO_{2}}}$;B.
Na₂CO₃(合理即可)
$\boldsymbol{{Na_{2}CO_{3}}}$。
2. 请写出图中涉及反应的化学方程式


答案: 1. A:$\boldsymbol{CO_{2}}$;B:$\boldsymbol{Na_{2}CO_{3}}$(合理即可)。
2. ①$\boldsymbol{HCl + NaOH = NaCl + H_{2}O}$、$\boldsymbol{H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O}$、$\boldsymbol{2HCl + Ca(OH)_{2} = CaCl_{2} + 2H_{2}O}$、$\boldsymbol{H_{2}SO_{4} + Ca(OH)_{2} = CaSO_{4} + 2H_{2}O}$;
②$\boldsymbol{2NaOH + MgCl_{2} = Mg(OH)_{2}\downarrow + 2NaCl}$、$\boldsymbol{2NaOH + CuSO_{4} = Cu(OH)_{2}\downarrow + Na_{2}SO_{4}}$、$\boldsymbol{Ca(OH)_{2} + MgCl_{2} = Mg(OH)_{2}\downarrow + CaCl_{2}}$、$\boldsymbol{Ca(OH)_{2} + CuSO_{4} = Cu(OH)_{2}\downarrow + CaSO_{4}}$;
③$\boldsymbol{2NaOH + CO_{2} = Na_{2}CO_{3} + H_{2}O}$、$\boldsymbol{2NaOH + SO_{2} = Na_{2}SO_{3} + H_{2}O}$、$\boldsymbol{Ca(OH)_{2} + CO_{2} = CaCO_{3}\downarrow + H_{2}O}$、$\boldsymbol{Ca(OH)_{2} + SO_{2} = CaSO_{3}\downarrow + H_{2}O}$;
④$\boldsymbol{Ca(OH)_{2} + Na_{2}CO_{3} = CaCO_{3}\downarrow + 2NaOH}$(工业制烧碱的原理)。

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