High School Chemistry

Chemical Reactions and Equations化学反应与方程式

Chemistry is the science of matter changing — iron rusting, fuels burning, salts dissolving to form precipitates. This guide teaches you to recognise when a chemical reaction has occurred, write and balance the equation that describes it, classify it by type (synthesis, decomposition, single replacement, double replacement, combustion), predict whether a precipitate forms using solubility rules, and write the net ionic equation that strips away spectator ions. Worked examples and KaTeX notation are used throughout.化学是研究物质转变的科学——铁生锈、燃料燃烧、盐溶解形成沉淀（沉淀，precipitate）。本指南教你识别化学反应（化学反应，chemical reaction）发生的证据，书写并配平（配平，balance）描述它的方程式（化学方程式，chemical equation），按类型分类（化合反应、分解反应、置换反应、燃烧），利用溶解度规则预测是否生成沉淀，并写出去除旁观离子后的净离子方程式（净离子方程式，net ionic equation）。全程使用例题与公式。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB IB Chemistry HL Reactivity 2 feederIB Chemistry HL Reactivity 2 衔接

Where this lives in your syllabus本单元在各大纲中的位置

HS-PS1-2 "Construct and revise an explanation for the outcome of a simple chemical reaction based on the outermost electron states of atoms, trends in the periodic table, and knowledge of the patterns of chemical properties. [Clarification Statement: Examples of chemical reactions could include the reaction of sodium and chlorine, of carbon and oxygen, or of carbon and hydrogen.] [Assessment Boundary: Assessment is limited to chemical reactions involving main group elements and combustion reactions.]" Maps to §3–§5 (reaction types, combustion) and §6–§7 (precipitation, net ionic)."根据原子的最外电子状态、元素周期表的趋势以及化学性质规律，构建并修正对简单化学反应结果的解释。[澄清说明：化学反应的示例包括钠与氯的反应、碳与氧的反应，或碳与氢的反应。] [评估边界：评估仅限于涉及主族元素的化学反应和燃烧反应。]" 对应 §3–§5（反应类型、燃烧）和 §6–§7（沉淀、净离子方程式）。
HS-PS1-7 "Use mathematical representations to support the claim that atoms, and therefore mass, are conserved during a chemical reaction. [Clarification Statement: Emphasis is on using mathematical ideas to communicate the proportional relationships between masses of atoms in the reactants and the products, and the translation of these relationships to the macroscopic scale using the mole as the conversion from the atomic to the macroscopic scale. Emphasis is on assessing students' use of mathematical thinking and not on memorization and rote application of problem-solving techniques.] [Assessment Boundary: Assessment does not include complex chemical reactions.]" Maps to §2 (balancing equations, conservation of mass)."用数学表示支持在化学反应中原子（因此质量）守恒的主张。[澄清说明：重点在于用数学思想传达反应物和产物中原子质量之间的比例关系，以及利用摩尔将这些关系从原子尺度转换到宏观尺度。重点在于评估学生的数学思维，而非对解题技巧的死记硬背。] [评估边界：评估不包括复杂的化学反应。]" 对应 §2（配平方程式，质量守恒）。

SNC2D Grade 10 Academic Science (Chemistry strand): "rearrangement of atoms in chemical reactions"; "law of conservation of mass" — the prerequisite balancing layer.SNC2D 10 年级学术理科（化学链）："化学反应中的原子重排"；"质量守恒定律"——先修配平层。
SCH3U Strand C (Chemical Reactions): Overall Expectation C3 "demonstrate an understanding of the different types of chemical reactions." Specific expectations: C2.2 "write balanced chemical equations to represent synthesis, decomposition, single displacement, double displacement, and combustion reactions, using the IUPAC nomenclature system"; C3.1 "identify various types of chemical reactions, including synthesis, decomposition, single displacement, double displacement, and combustion"; C3.2 "explain the difference between a complete combustion reaction and an incomplete combustion reaction"; C2.5 "predict the products of single displacement reactions, using the metal activity series and the halogen series"; C2.6 "predict the products of double displacement reactions (e.g., the formation of precipitates or gases; neutralization)."SCH3U C 单元（化学反应）：总体期望 C3"展示对不同类型化学反应的理解"。具体期望：C2.2"用 IUPAC 命名系统书写表示化合、分解、单置换、双置换和燃烧反应的配平化学方程式"；C3.1"识别各种类型的化学反应，包括化合、分解、单置换、双置换和燃烧"；C3.2"解释完全燃烧反应与不完全燃烧反应的区别"；C2.5"利用金属活动序和卤素序预测单置换反应的产物"；C2.6"预测双置换反应的产物（例如，沉淀或气体的形成；中和）"。

Science 10 Big Idea: "Energy change is required as atoms rearrange in chemical processes." Content: "rearrangement of atoms in chemical reactions"; "law of conservation of mass"; "energy change during chemical reactions" — foundational reaction layer for Chemistry 11.Science 10 大概念："化学过程中原子重排需要能量变化。"内容："化学反应中的原子重排"；"质量守恒定律"；"化学反应中的能量变化"——Chemistry 11 的基础反应层。
Chemistry 11 Big Idea: "Matter and energy are conserved in chemical reactions." Content: "reactions" — elaborated as "predicting products, reactants and energy changes (ΔH)"; "stoichiometric calculations: — mass — number of molecules — gas volumes — molar quantities — excess and limiting reactants"; "green chemistry: development of sustainable processes and technologies that reduce negative impacts on the environment."Chemistry 11 大概念："物质和能量在化学反应中守恒。"内容："反应"——细化为"预测产物、反应物与能量变化（ΔH）"；"化学计量计算：质量 · 分子数 · 气体体积 · 摩尔量 · 过量与限量试剂"；"绿色化学：开发减少对环境负面影响的可持续过程和技术"。

Science 10 Unit A (Energy and Matter in Chemical Change): prerequisite layer — conservation of mass and energy in chemical changes, foundational reaction types.Science 10 A 单元（化学变化中的能量与物质）：先修层——化学变化中的质量与能量守恒，基础反应类型。
Chemistry 20 Unit D "Quantitative Relationships in Chemical Changes", General Outcome 1: "explain how balanced chemical equations indicate the quantitative relationships between reactants and products involved in chemical changes." Key Concepts: "chemical reaction equations · net ionic equations · spectator ions · reaction stoichiometry · precipitation." Knowledge outcomes: "predict the product(s) of a chemical reaction based upon the reaction type"; "recall the balancing of chemical equations in terms of atoms, molecules and moles"; "write balanced ionic and net ionic equations, including identification of spectator ions, for reactions taking place in aqueous solutions."Chemistry 20 D 单元"化学变化中的定量关系"，总目标 1："解释配平的化学方程式如何表明化学变化中反应物和产物之间的定量关系"。关键概念："化学反应方程式 · 净离子方程式 · 旁观离子 · 反应化学计量 · 沉淀"。知识结果："根据反应类型预测化学反应的产物"；"回顾以原子、分子和摩尔为单位配平化学方程式"；"书写水溶液中反应的配平离子和净离子方程式，包括识别旁观离子"。

Read me first先读这里

How to use this guide如何使用本指南

Chemical reactions and equations is a core topic in every curriculum we map to, and the four curricula agree on the essential scope: evidence of reaction, balancing equations by atom conservation, the five reaction types (synthesis, decomposition, single replacement, double replacement, combustion), solubility rules, and net ionic equations. Depth differences are minor. US NGSS (HS-PS1-2, HS-PS1-7) limits assessed reactions to main-group and combustion; Ontario SCH3U Strand C covers all five types explicitly with predict-products expectations; BC Chemistry 11 frames reactions through conservation and stoichiometry; Alberta Chemistry 20 Unit D adds net ionic equations and spectator ions. All sections here are relevant to every curriculum. The table below shows the alignment.化学反应与方程式是我们所对照的所有大纲的核心主题，四套大纲在基本范围上一致：反应的证据、原子守恒配平方程式、五种反应类型（化合、分解、单置换、双置换、燃烧）、溶解度规则与净离子方程式。深度差异较小。US NGSS（HS-PS1-2、HS-PS1-7）将评估反应限于主族元素和燃烧；安大略 SCH3U C 单元明确涵盖所有五种类型并有预测产物要求；BC Chemistry 11 通过守恒和化学计量学框架反应；阿尔伯塔 Chemistry 20 D 单元加入净离子方程式和旁观离子。本指南所有节对每套大纲都有意义。下表显示对照关系。

If you are in…如果你在…	Focus on these sections重点学习	Notes说明	Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学	§1–§5 (evidence, balancing, synthesis, decomposition, single/double replacement, combustion) — core under HS-PS1-2 and HS-PS1-7. NGSS limits assessed reactions to main-group elements and combustion.§1–§5（证据、配平、化合、分解、单/双置换、燃烧）——HS-PS1-2 与 HS-PS1-7 下的核心。NGSS 将评估反应限于主族元素和燃烧。	§6–§7 (solubility rules, net ionic equations): valuable context but no dedicated NGSS PE; approached through HS-PS1-2 reaction outcomes.§6–§7（溶解度规则、净离子方程式）：有价值的背景，但无专门的 NGSS PE；通过 HS-PS1-2 反应结果来处理。	`NGSS HS-PS1 (Chemistry)` — HS-PS1-2 and HS-PS1-7 PE + Clarification + Assessment Boundary— HS-PS1-2 与 HS-PS1-7 表现期望 + 澄清 + 评估边界
🇨🇦 ON Grade 10/11 — SNC2D / SCH3U安大略 10/11 年级 — SNC2D / SCH3U	All seven sections. SCH3U C2.2 explicitly lists all five reaction types; C2.5 predict single-displacement using activity series; C2.6 predict double-displacement precipitates; Strand E (solutions) introduces net ionic equations and solubility.全部七节。SCH3U C2.2 明确列出所有五种反应类型；C2.5 利用活动序预测单置换；C2.6 预测双置换沉淀；E 单元（溶液）引入净离子方程式和溶解度。	Net ionic equations formally enter in SCH3U Strand E2.5 ("write balanced net ionic equations to represent precipitation and neutralization reactions") — covered in §7 here.净离子方程式正式出现在 SCH3U E2.5（"书写配平的净离子方程式以表示沉淀和中和反应"）——此处在 §7 中涵盖。	`Ontario SCH3U/4U Chemistry` — SCH3U Strand C Overall Expectations C1–C3, Specific Expectations C2.2, C2.4–C2.6, C3.1–C3.3; Strand E2.5— SCH3U C 单元总体期望 C1–C3，具体期望 C2.2、C2.4–C2.6、C3.1–C3.3；E2.5
🇨🇦 BC Science 10 / Chemistry 11BC Science 10 / Chemistry 11	All seven sections. Chemistry 11 Big Idea "Matter and energy are conserved in chemical reactions"; Content "reactions: predicting products, reactants and energy changes (ΔH)." Science 10 lays the prerequisite conservation of mass and atom rearrangement.全部七节。Chemistry 11 大概念"物质和能量在化学反应中守恒"；内容"反应：预测产物、反应物与能量变化（ΔH）"。Science 10 奠定质量守恒和原子重排的先修知识。	BC Chemistry 11 integrates solubility into its solutions content ("solubility: dissociation of ions, dissociation equation") — §6 is directly relevant.BC Chemistry 11 将溶解度整合到溶液内容中（"溶解度：离子解离，解离方程式"）——§6 直接相关。	`BC Chemistry 11/12` — Chemistry 11 Big Idea "Matter and energy are conserved in chemical reactions"; Content "reactions"; Science 10 Content "rearrangement of atoms in chemical reactions"; "law of conservation of mass"— Chemistry 11 大概念"物质和能量在化学反应中守恒"；内容"反应"；Science 10 内容"化学反应中的原子重排"；"质量守恒定律"
🇨🇦 AB Science 10 / Chemistry 20阿尔伯塔 Science 10 / Chemistry 20	All seven sections. Chemistry 20 Unit D GO1 requires: "predict the product(s) of a chemical reaction based upon the reaction type"; "recall the balancing of chemical equations in terms of atoms, molecules and moles"; "write balanced ionic and net ionic equations, including identification of spectator ions." Net ionic equations are explicitly required.全部七节。Chemistry 20 D 单元 GO1 要求："根据反应类型预测化学反应的产物"；"回顾以原子、分子和摩尔为单位配平化学方程式"；"书写配平的离子和净离子方程式，包括识别旁观离子"。净离子方程式是明确要求的。	Alberta places strong-acid/strong-base titration in Chemistry 20 Unit D GO2 — introduced in §7 but fully covered in Unit 9 (Acids, Bases & pH).阿尔伯塔将强酸/强碱滴定放在 Chemistry 20 D 单元 GO2——在 §7 中简介，在第 9 单元（酸、碱与 pH）中完整涵盖。	`Alberta Chemistry 20/30` — Chemistry 20 Unit D General Outcomes GO1/GO2, Key Concepts, knowledge outcome text; Science 10 Unit A prerequisite note— Chemistry 20 D 单元总目标 GO1/GO2，关键概念，知识结果文本；Science 10 A 单元先修说明
🇺🇸 IB / AP feeder trackIB / AP 衔接轨道	All seven sections. IB Chemistry HL Reactivity 2 ("How Much, How Fast, How Far?") and AP Chemistry Unit 4 (Chemical Reactions) assume fluent reaction classification, balancing, net ionic equations, and solubility rules from the first week.全部七节。IB Chemistry HL Reactivity 2（"多少？多快？多远？"）与 AP Chemistry Unit 4（化学反应）从第一周起就默认你熟练反应分类、配平、净离子方程式和溶解度规则。	Nothing — this unit is assumed prerequisite for all higher-level reaction mechanism, kinetics, and equilibrium content.无 — 本单元是所有高级反应机理、动力学和平衡内容的先修前提。	`NGSS HS-PS1 (Chemistry)` — see "What This Feeds Into" for IB Chemistry HL Reactivity 2 link— 见"本单元的去向"中的 IB Chemistry HL Reactivity 2 链接

Once you have located your row, use the two cards below for the pace at which you should work.找到所在行后，用下面两张卡片决定推进速度。

If you are cramming the night before如果你在临阵磨枪

Memorise five things: (1) a balanced equation has equal atom counts on both sides; (2) synthesis: A + B → AB; decomposition: AB → A + B; (3) in single replacement, a more active metal displaces a less active one; (4) in double replacement, swap the cations of two ionic compounds; (5) combustion of a hydrocarbon gives CO$_2$ and H$_2$O. Read every cram-cheat box. Confirm your solubility table rules (§6) before any precipitation question.背熟五件事：(1) 配平方程式两侧的原子数相等；(2) 化合：A + B → AB；分解：AB → A + B；(3) 单置换中，活动性较强的金属置换活动性较弱的金属；(4) 双置换中，交换两种离子化合物的阳离子；(5) 烃的燃烧产生 CO$_2$ 和 H$_2$O。读每个速记框。在解答任何沉淀问题之前确认溶解度表规则（§6）。

If you are going for the top mark如果你目标顶分

For every equation: identify the reaction type first, then predict products before balancing. Practice net ionic equations by writing the full molecular equation, splitting strong electrolytes into ions, then cancelling spectator ions. Know the solubility rules cold so you can instantly judge whether a precipitate forms. For combustion, balance oxygen last. For single replacement, use the metal activity series to confirm the reaction is spontaneous. ON SCH3U C2.5, AB Chem 20 D GO1, and IB Reactivity 2 all assess predict-then-balance skill, not just copying a given equation.对每个方程式：先确定反应类型，然后在配平前预测产物。通过书写完整的分子方程式、将强电解质拆分为离子，然后消去旁观离子来练习净离子方程式。熟记溶解度规则，以便立即判断是否生成沉淀。对于燃烧，最后配平氧气。对于单置换，使用金属活动序确认反应是自发的。ON SCH3U C2.5、AB Chem 20 D GO1 和 IB Reactivity 2 都评估预测然后配平的技能，而不仅仅是抄写已给出的方程式。

Section 1 · Evidence of a chemical reaction第 1 节 · 化学反应的证据

Evidence of a Chemical Reaction化学反应的证据

Five observable clues that a new substance has formed.五种可观测的迹象表明新物质已生成。

Color change颜色变化 — e.g. copper wire turns green in moist air (Cu → Cu$_2$(OH)$_2$CO$_3$); iron rusts orange-brown.— 例如铜线在潮湿空气中变绿（Cu → Cu$_2$(OH)$_2$CO$_3$）；铁生成橙棕色铁锈。
Gas produced产生气体 — bubbles form that were not present before (e.g. acid + carbonate → CO$_2$).— 出现之前不存在的气泡（例如酸 + 碳酸盐 → CO$_2$）。
Precipitate formed生成沉淀 — an insoluble solid appears in solution (e.g. mixing Pb(NO$_3$)$_2$ and KI gives yellow PbI$_2$).— 溶液中出现不溶固体（例如混合 Pb(NO$_3$)$_2$ 和 KI 生成黄色 PbI$_2$）。
Energy change能量变化 — the container becomes noticeably hot (exothermic) or cold (endothermic).— 容器变得明显发热（放热）或变冷（吸热）。
Light emitted发光 — e.g. a sparkler burning or chemiluminescence in glow sticks.— 例如烟花棒燃烧或荧光棒中的化学发光。

Physical vs chemical change:物理变化与化学变化： A physical change alters form but not composition (melting ice, tearing paper). A chemical change (reaction) produces one or more new substances with different chemical formulas — the evidence above signals a new substance. NGSS HS-PS1-2 requires constructing explanations for "the outcome of a simple chemical reaction." Ontario SCH3U C2.1 lists "precipitate, acidic, and basic" as assessed terminology.物理变化改变形态但不改变成分（冰融化、撕纸）。化学变化（反应）产生具有不同化学式的新物质——上述证据表明有新物质生成。NGSS HS-PS1-2 要求构建对"简单化学反应结果"的解释。安大略 SCH3U C2.1 把"沉淀、酸性、碱性"列为被评估的术语。

Worked Example 1 · Identifying evidence of reaction例题 1 · 识别化学反应的证据

A student mixes aqueous sodium carbonate (Na$_2$CO$_3$) with dilute hydrochloric acid (HCl). She observes vigorous bubbling and the solution warms slightly. (a) Identify two pieces of evidence that a chemical reaction occurred. (b) Name the gas produced.学生将碳酸钠（Na$_2$CO$_3$）水溶液与稀盐酸（HCl）混合。她观察到剧烈冒泡，溶液略微变热。(a) 指出两条发生化学反应的证据。(b) 命名产生的气体。

(a) Evidence 1:(a) 证据 1： Gas bubbles are produced (a gas was not present in the reactants — CO$_2$ forms as a new substance).产生气泡（反应物中不存在气体——CO$_2$ 作为新物质生成）。

Evidence 2:证据 2： The solution warms (exothermic energy change — heat is released to the surroundings).溶液变热（放热能量变化——热量释放到周围环境）。

(b) The gas is carbon dioxide (CO$_2$).(b) 气体是二氧化碳（CO$_2$）。 The reaction is: Na$_2$CO$_3$(aq) + 2 HCl(aq) → 2 NaCl(aq) + H$_2$O(l) + CO$_2$(g).反应方程式为：Na$_2$CO$_3$(aq) + 2 HCl(aq) → 2 NaCl(aq) + H$_2$O(l) + CO$_2$(g)。

Which observation is the strongest evidence that a chemical reaction has occurred?下列哪种观察是化学反应发生的最强证据？

§1 · Q1

A solid dissolves in water固体溶解在水中

The temperature drops when ice is added加入冰后温度下降

A yellow solid forms when two colourless solutions are mixed两种无色溶液混合后形成黄色固体

Water is heated to form steam水被加热形成蒸汽

A precipitate forming — a new insoluble solid appearing — is clear evidence that new substances with different chemical formulas have been produced. Dissolving, cooling by added ice, and boiling are physical changes (same chemical composition before and after).沉淀形成——出现新的不溶固体——是产生了具有不同化学式的新物质的明确证据。溶解、加冰冷却和沸腾是物理变化（前后化学成分相同）。

Dissolving, temperature change from adding ice, and boiling are all physical changes — the substance's chemical identity does not change. A precipitate forming signals a new chemical compound.溶解、加冰引起的温度变化和沸腾都是物理变化——物质的化学本质不变。沉淀形成表明有新的化合物生成。

Iron rusting is classified as a chemical change because:铁生锈被归类为化学变化，因为：

§1 · Q2

The iron changes shape铁的形状改变了

A new substance (iron oxide, Fe$_2$O$_3$) with a different chemical formula is produced产生了具有不同化学式的新物质（氧化铁，Fe$_2$O$_3$）

The mass of the iron decreases铁的质量减少了

The colour changes from silver to orange颜色从银色变为橙色

The defining criterion for a chemical change is the formation of a new substance with a different chemical formula. Fe$_2$O$_3$ has a different formula from Fe — it is a new compound. Colour change alone is insufficient (it can accompany physical changes); shape change and mass change are incidental.化学变化的决定性标准是生成具有不同化学式的新物质。Fe$_2$O$_3$ 的化学式与 Fe 不同——它是一种新化合物。仅颜色变化不足以判断（物理变化也可能伴随颜色变化）；形状变化和质量变化是附带的。

Colour change alone, shape change, or mass change can accompany physical changes. The key is whether new chemical formulas appear. Rusting converts Fe into Fe$_2$O$_3$ — a chemically distinct compound.仅颜色变化、形状变化或质量变化都可能伴随物理变化。关键是是否出现新的化学式。生锈将 Fe 转变为 Fe$_2$O$_3$——一种化学上截然不同的化合物。

Section 2 · Balancing chemical equations第 2 节 · 配平化学方程式

Balancing Chemical Equations配平化学方程式

Conservation of mass: atoms are neither created nor destroyed.质量守恒：原子既不会被创造也不会被消灭。

Law of conservation of mass质量守恒定律 — in a chemical reaction atoms are rearranged, not created or destroyed. A balanced equation has equal numbers of each type of atom on both sides. NGSS HS-PS1-7 and BC Science 10 both cite this law directly.— 在化学反应中，原子被重新排列，而不是被创造或消灭。配平的方程式两侧每种原子的数量相等。NGSS HS-PS1-7 与 BC Science 10 均直接引用此定律。
Balancing rules配平规则
- Change coefficients (numbers in front of formulas); never change subscripts.改变系数（化学式前面的数字）；永远不要改变下标。
- Balance one element at a time; leave O and H for last if possible.每次配平一种元素；如果可能，最后配平 O 和 H。
- Coefficients must be the lowest whole-number ratio (simplify if needed).系数必须是最低整数比（如需要则化简）。
State symbols状态符号 — (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water). Ontario SCH3U C2.2 and AB Chemistry 20 D expect state symbols in balanced equations.— (s) 固体，(l) 液体，(g) 气体，(aq) 水溶液（溶于水）。安大略 SCH3U C2.2 与 AB Chemistry 20 D 要求在配平方程式中注明状态符号。

Worked Example 2 · Balancing by inspection例题 2 · 观察法配平

Balance the equation for the reaction of iron with oxygen to form iron(III) oxide:配平铁与氧气反应生成氧化铁（III）的方程式：

Fe(s) + O$_2$(g) → Fe$_2$O$_3$(s)Fe(s) + O$_2$(g) → Fe$_2$O$_3$(s)

Step 1: Balance Fe.第 1 步：配平 Fe。 There are 2 Fe in Fe$_2$O$_3$, so put a 4 in front of Fe and a 2 in front of Fe$_2$O$_3$ (to get 4 Fe on each side): 4 Fe + O$_2$ → 2 Fe$_2$O$_3$.Fe$_2$O$_3$ 中有 2 个 Fe，所以在 Fe 前放 4，在 Fe$_2$O$_3$ 前放 2（使两侧各有 4 个 Fe）：4 Fe + O$_2$ → 2 Fe$_2$O$_3$。

Step 2: Balance O.第 2 步：配平 O。 Right side: $2 \times 3 = 6$ O atoms. O$_2$ supplies 2 O per molecule, so we need 3 molecules: 4 Fe + 3 O$_2$ → 2 Fe$_2$O$_3$.右侧：$2 \times 3 = 6$ 个 O 原子。O$_2$ 每分子提供 2 个 O，故需要 3 个分子：4 Fe + 3 O$_2$ → 2 Fe$_2$O$_3$。

Check:核验： Left: 4 Fe, 6 O. Right: 4 Fe, 6 O. ✓ Lowest-whole-number coefficients: 4, 3, 2.左侧：4 Fe，6 O。右侧：4 Fe，6 O。✓ 最低整数系数：4、3、2。

$$ 4\,\text{Fe(s)} + 3\,\text{O}_2\text{(g)} \rightarrow 2\,\text{Fe}_2\text{O}_3\text{(s)} $$

Which is the correctly balanced equation for the reaction of hydrogen gas with oxygen gas to form water?下列哪个是氢气与氧气反应生成水的正确配平方程式？

§2 · Q1

H$_2$(g) + O$_2$(g) → H$_2$O(l)H$_2$(g) + O$_2$(g) → H$_2$O(l)

H$_2$(g) + O(g) → H$_2$O(l)H$_2$(g) + O(g) → H$_2$O(l)

2 H$_2$(g) + O$_2$(g) → H$_4$O$_2$(l)2 H$_2$(g) + O$_2$(g) → H$_4$O$_2$(l)

2 H$_2$(g) + O$_2$(g) → 2 H$_2$O(l)2 H$_2$(g) + O$_2$(g) → 2 H$_2$O(l)

Left: 4 H (from 2 H$_2$) and 2 O (from O$_2$). Right: 4 H and 2 O (from 2 H$_2$O). Balanced. Note: H$_4$O$_2$ is not a real formula — you cannot change subscripts to balance.左侧：4 H（来自 2 H$_2$）和 2 O（来自 O$_2$）。右侧：4 H 和 2 O（来自 2 H$_2$O）。已配平。注：H$_4$O$_2$ 不是真实的化学式——不能通过改变下标来配平。

Never change subscripts. Use only coefficients. Option A has 2 O on left, 1 O on right — unbalanced. Option B uses atomic O, not O$_2$. Option C invents a formula. Option D is the correct balance.永远不要改变下标。只使用系数。选项 A 左侧有 2 O，右侧有 1 O——未配平。选项 B 使用原子 O 而不是 O$_2$。选项 C 发明了一个化学式。选项 D 是正确的配平。

In balancing a chemical equation, which action is permitted?在配平化学方程式时，以下哪个操作是允许的？

§2 · Q2

Changing the coefficient in front of a formula改变化学式前面的系数

Changing the subscript inside a formula改变化学式内部的下标

Adding new atoms not in the original reactants添加原反应物中没有的新原子

Removing atoms from a product to make it balance从产物中删除原子使其配平

Only coefficients (the whole numbers in front of formulas) may be changed. Changing subscripts alters the identity of the compound. Adding or removing atoms violates conservation of mass.只能改变系数（化学式前面的整数）。改变下标会改变化合物的本质。添加或删除原子违反质量守恒定律。

Subscripts define which compound you have — changing them creates a different substance. Balancing is done only by adjusting coefficients (the prefixes) so atom counts match on both sides.下标定义了化合物的身份——改变它们会产生不同的物质。配平只通过调整系数（前缀）来实现，使两侧的原子数相匹配。

Section 3 · Synthesis and decomposition第 3 节 · 化合反应与分解反应

Synthesis and Decomposition Reactions化合反应与分解反应

Two reactants combine, or one reactant breaks apart.两种反应物结合，或一种反应物分解。

Synthesis化合反应 (Synthesis) — general form: A + B → AB. Two or more substances combine to form a single, more complex product.— 通式：A + B → AB。两种或更多物质结合形成一种更复杂的产物。
- Metal + O$_2$ → metal oxide: $4\,\text{Na(s)} + \text{O}_2\text{(g)} \to 2\,\text{Na}_2\text{O(s)}$金属 + O$_2$ → 金属氧化物：$4\,\text{Na(s)} + \text{O}_2\text{(g)} \to 2\,\text{Na}_2\text{O(s)}$
- Non-metal + O$_2$ → non-metal oxide: $\text{S(s)} + \text{O}_2\text{(g)} \to \text{SO}_2\text{(g)}$非金属 + O$_2$ → 非金属氧化物：$\text{S(s)} + \text{O}_2\text{(g)} \to \text{SO}_2\text{(g)}$
- Metal oxide + H$_2$O → metal hydroxide: $\text{CaO(s)} + \text{H}_2\text{O(l)} \to \text{Ca(OH)}_2\text{(aq)}$金属氧化物 + H$_2$O → 金属氢氧化物：$\text{CaO(s)} + \text{H}_2\text{O(l)} \to \text{Ca(OH)}_2\text{(aq)}$
Decomposition分解反应 (Decomposition) — general form: AB → A + B. A single compound breaks down into two or more simpler substances. Often needs energy input (heat, light, or electricity).— 通式：AB → A + B。一种化合物分解为两种或更多更简单的物质。通常需要能量输入（热、光或电）。
- Thermal decomposition of carbonate: $\text{CaCO}_3\text{(s)} \xrightarrow{\Delta} \text{CaO(s)} + \text{CO}_2\text{(g)}$碳酸盐的热分解：$\text{CaCO}_3\text{(s)} \xrightarrow{\Delta} \text{CaO(s)} + \text{CO}_2\text{(g)}$
- Electrolysis of water: $2\,\text{H}_2\text{O(l)} \xrightarrow{\text{elec.}} 2\,\text{H}_2\text{(g)} + \text{O}_2\text{(g)}$水的电解：$2\,\text{H}_2\text{O(l)} \xrightarrow{\text{elec.}} 2\,\text{H}_2\text{(g)} + \text{O}_2\text{(g)}$

SCH3U C3.1 explicitly names synthesis and decomposition. NGSS HS-PS1-2 uses "carbon and oxygen" (synthesis) as an example. AB Chemistry 20 D GO1 requires predicting products by reaction type.SCH3U C3.1 明确列出化合和分解反应。NGSS HS-PS1-2 使用"碳与氧"（化合）作为示例。AB Chemistry 20 D GO1 要求根据反应类型预测产物。

Worked Example 3 · Predicting the product of a synthesis reaction例题 3 · 预测化合反应的产物

Magnesium ribbon burns in oxygen. (a) Identify the reaction type. (b) Write the balanced equation. (c) Describe one observable piece of evidence.镁带在氧气中燃烧。(a) 确定反应类型。(b) 书写配平方程式。(c) 描述一条可观测的证据。

(a) Reaction type: synthesis.(a) 反应类型：化合反应。 Two elements (Mg and O$_2$) combine to form one compound (MgO) — the A + B → AB pattern.两种元素（Mg 和 O$_2$）结合形成一种化合物（MgO）——A + B → AB 的模式。

(b) Balanced equation:(b) 配平方程式：

$$ 2\,\text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\,\text{MgO(s)} $$

Check: 2 Mg left and right; 2 O left and right. ✓核验：左右各 2 Mg；左右各 2 O。✓

(c) Evidence:(c) 证据： Brilliant white flame / bright white light emitted; a white powder (MgO) remains — both a color change and light emission indicating a chemical reaction.发出耀眼的白色火焰/强白光；留下白色粉末（MgO）——颜色变化和光的发射均表明发生了化学反应。

Which equation represents a decomposition reaction?下列哪个方程式表示分解反应？

§3 · Q1

2 H$_2$O$_2$(aq) → 2 H$_2$O(l) + O$_2$(g)2 H$_2$O$_2$(aq) → 2 H$_2$O(l) + O$_2$(g)

2 Na(s) + Cl$_2$(g) → 2 NaCl(s)2 Na(s) + Cl$_2$(g) → 2 NaCl(s)

Zn(s) + 2 HCl(aq) → ZnCl$_2$(aq) + H$_2$(g)Zn(s) + 2 HCl(aq) → ZnCl$_2$(aq) + H$_2$(g)

CH$_4$(g) + 2 O$_2$(g) → CO$_2$(g) + 2 H$_2$O(g)CH$_4$(g) + 2 O$_2$(g) → CO$_2$(g) + 2 H$_2$O(g)

One reactant (H$_2$O$_2$) breaks down into two products (H$_2$O and O$_2$) — the AB → A + B pattern. This is the decomposition of hydrogen peroxide, catalysed by MnO$_2$ in labs. Option B is synthesis; C is single replacement; D is combustion.一种反应物（H$_2$O$_2$）分解为两种产物（H$_2$O 和 O$_2$）——AB → A + B 的模式。这是过氧化氢的分解，实验室中由 MnO$_2$ 催化。选项 B 是化合；C 是单置换；D 是燃烧。

Decomposition: one reactant → multiple products. Option A has one reactant (H$_2$O$_2$) breaking into two products. Options B, C, D all have multiple reactants or a different pattern.分解反应：一种反应物 → 多种产物。选项 A 有一种反应物（H$_2$O$_2$）分解为两种产物。选项 B、C、D 都有多种反应物或不同的模式。

Sulfur trioxide (SO$_3$) is formed by the synthesis reaction of sulfur dioxide and oxygen. Which balanced equation correctly represents this?三氧化硫（SO$_3$）由二氧化硫和氧气的化合反应生成。下列哪个配平方程式正确表示了这一过程？

§3 · Q2

SO$_2$(g) + O$_2$(g) → SO$_3$(g)SO$_2$(g) + O$_2$(g) → SO$_3$(g)

SO$_2$(g) + O(g) → SO$_3$(g)SO$_2$(g) + O(g) → SO$_3$(g)

2 SO$_2$(g) + O$_2$(g) → 2 SO$_4$(g)2 SO$_2$(g) + O$_2$(g) → 2 SO$_4$(g)

2 SO$_2$(g) + O$_2$(g) → 2 SO$_3$(g)2 SO$_2$(g) + O$_2$(g) → 2 SO$_3$(g)

Left: 2 S, 4 O (from 2 SO$_2$) + 2 O (from O$_2$) = 6 O. Right: 2 S, 6 O (from 2 SO$_3$). Balanced. Option A gives 1 S, 4 O left vs 1 S, 3 O right — unbalanced. Option B uses atomic O. Option C invents SO$_4$ as a product (impossible from SO$_2$ + O$_2$).左侧：2 S，4 O（来自 2 SO$_2$）+ 2 O（来自 O$_2$）= 6 O。右侧：2 S，6 O（来自 2 SO$_3$）。已配平。选项 A 给出左侧 1 S、4 O 与右侧 1 S、3 O——未配平。选项 B 使用原子 O。选项 C 将 SO$_4$ 作为产物（从 SO$_2$ + O$_2$ 不可能得到）。

Balance atoms: 2 SO$_2$ gives 2 S and 4 O on the left; O$_2$ adds 2 more O (total 6 O). The product 2 SO$_3$ gives 2 S and 6 O on the right. Check with option D.配平原子：2 SO$_2$ 在左侧给出 2 S 和 4 O；O$_2$ 再加 2 O（共 6 O）。产物 2 SO$_3$ 在右侧给出 2 S 和 6 O。选项 D 正确。

Section 4 · Single and double replacement第 4 节 · 单置换与双置换

Single and Double Replacement Reactions单置换与双置换反应

One element (or ion pair) swaps places with another.一种元素（或离子对）与另一种交换位置。

Single replacement单置换反应 (Single replacement) — general form: A + BC → AC + B. A more active element displaces a less active one from its compound.— 通式：A + BC → AC + B。活动性较强的元素从其化合物中置换出活动性较弱的元素。
- Metal displacement: $\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \to \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$ — Zn is above Cu in the activity series.金属置换：$\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \to \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$ — Zn 在活动序中位于 Cu 之上。
- Metal in acid: $\text{Fe(s)} + 2\,\text{HCl(aq)} \to \text{FeCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ — metals above H in the series react with dilute acids.金属与酸：$\text{Fe(s)} + 2\,\text{HCl(aq)} \to \text{FeCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ — 活动序中 H 以上的金属与稀酸反应。
- Rule: no reaction if the free element is less active than the one it would displace. E.g. Cu + ZnSO$_4$ → no reaction (Cu is below Zn in the series).规则：如果自由元素的活动性低于它将置换的元素，则无反应。例如 Cu + ZnSO$_4$ → 无反应（Cu 在活动序中位于 Zn 之下）。
Double replacement双置换反应 (Double replacement) — general form: AB + CD → AD + CB. Cations and anions of two ionic compounds switch partners. A reaction occurs if a precipitate, gas, or water is produced. SCH3U C2.6 and AB Chem 20 D GO1 both assess predicting double-replacement products.— 通式：AB + CD → AD + CB。两种离子化合物的阳离子和阴离子交换伙伴。如果产生沉淀、气体或水，则发生反应。SCH3U C2.6 与 AB Chem 20 D GO1 均考查预测双置换反应产物。

Worked Example 4 · Single replacement using the activity series例题 4 · 利用活动序的单置换反应

Will a reaction occur when a strip of copper (Cu) is placed in silver nitrate solution (AgNO$_3$(aq))? If so, write the balanced equation. (Activity series excerpt, most to least active: Mg, Al, Zn, Fe, Ni, Cu, Ag, Au.)将铜片（Cu）放入硝酸银溶液（AgNO$_3$(aq)）中时，会发生反应吗？如果是，写出配平方程式。（活动序摘录，从最活泼到最不活泼：Mg、Al、Zn、Fe、Ni、Cu、Ag、Au。）

Activity check:活动性检验： Cu is above Ag in the activity series, so Cu can displace Ag from its solution. Reaction proceeds.Cu 在活动序中位于 Ag 之上，故 Cu 能从溶液中置换出 Ag。反应进行。

Predict products:预测产物： Cu replaces Ag; Ag deposits as solid; Cu$^{2+}$ enters solution with NO$_3^-$.Cu 置换 Ag；Ag 以固体析出；Cu$^{2+}$ 以 NO$_3^-$ 进入溶液。

$$ \text{Cu(s)} + 2\,\text{AgNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\,\text{Ag(s)} $$

Check:核验： 1 Cu each side; 2 Ag each side; 2 NO$_3$ each side. ✓ Observable evidence: silvery solid deposits on the copper; solution turns blue (Cu$^{2+}$(aq)).两侧各 1 Cu；两侧各 2 Ag；两侧各 2 NO$_3$。✓ 可观测证据：铜上析出银白色固体；溶液变蓝（Cu$^{2+}$(aq)）。

Magnesium is placed in copper(II) sulfate solution. Which statement correctly describes what happens? (Activity series: Mg is above Cu.)将镁放入硫酸铜（II）溶液中。下列哪个说法正确描述了发生的情况？（活动序：Mg 在 Cu 之上。）

§4 · Q1

No reaction occurs because copper is a metal不发生反应，因为铜是金属

Copper displaces magnesium from the solid铜从固体中置换出镁

Magnesium displaces copper; a red-brown solid forms and the blue colour fades镁置换铜；形成红棕色固体，蓝色褪去

Magnesium dissolves without any other change镁溶解但没有其他变化

Mg is higher in the activity series than Cu, so Mg displaces Cu$^{2+}$ from solution. The equation is Mg(s) + CuSO$_4$(aq) → MgSO$_4$(aq) + Cu(s). Cu deposits as a red-brown solid; the blue Cu$^{2+}$ colour fades as Mg$^{2+}$ (colourless) forms.Mg 在活动序中高于 Cu，故 Mg 从溶液中置换出 Cu$^{2+}$。方程式为 Mg(s) + CuSO$_4$(aq) → MgSO$_4$(aq) + Cu(s)。Cu 以红棕色固体析出；蓝色 Cu$^{2+}$ 褪去，因为 Mg$^{2+}$（无色）形成。

For single replacement: the element higher in the activity series displaces the one lower. Mg is more active than Cu, so Mg replaces Cu, not the other way. A reaction definitely occurs.对于单置换：活动序中较高的元素置换较低的元素。Mg 比 Cu 活泼，故 Mg 置换 Cu，而非相反。反应肯定会发生。

Aqueous solutions of barium chloride (BaCl$_2$) and sodium sulfate (Na$_2$SO$_4$) are mixed. Which type of reaction is this and what is the insoluble product?氯化钡（BaCl$_2$）和硫酸钠（Na$_2$SO$_4$）的水溶液混合。这是哪种类型的反应，不溶性产物是什么？

§4 · Q2

Single replacement; BaSO$_4$ precipitates单置换；BaSO$_4$ 沉淀

Double replacement; BaSO$_4$ precipitates双置换；BaSO$_4$ 沉淀

Synthesis; NaCl is produced as a solid化合；NaCl 以固体生成

Decomposition; barium and sulfur are released分解；钡和硫被释放

BaCl$_2$ + Na$_2$SO$_4$ → BaSO$_4$(s) + 2 NaCl(aq). The cations and anions swap partners (Ba$^{2+}$ pairs with SO$_4^{2-}$; Na$^+$ pairs with Cl$^-$) — double replacement. BaSO$_4$ is insoluble (white precipitate). NaCl remains in solution.BaCl$_2$ + Na$_2$SO$_4$ → BaSO$_4$(s) + 2 NaCl(aq)。阳离子和阴离子交换伙伴（Ba$^{2+}$ 与 SO$_4^{2-}$ 配对；Na$^+$ 与 Cl$^-$ 配对）——双置换。BaSO$_4$ 不溶（白色沉淀）。NaCl 留在溶液中。

No element replaces another in the compound (that would be single replacement). Two ionic compounds exchange anion partners — the AB + CD → AD + CB pattern. BaSO$_4$ is a well-known insoluble sulfate and precipitates.没有元素在化合物中置换另一个（那将是单置换）。两种离子化合物交换阴离子伙伴——AB + CD → AD + CB 的模式。BaSO$_4$ 是众所周知的不溶性硫酸盐，会沉淀。

Section 5 · Combustion第 5 节 · 燃烧

Combustion Reactions燃烧反应

Fuel + O$_2$ → oxides (+ energy).燃料 + O$_2$ → 氧化物（+ 能量）。

Complete combustion of a hydrocarbon烃的完全燃烧 — requires excess O$_2$. Products: CO$_2$(g) and H$_2$O(g). General form:— 需要过量 O$_2$。产物：CO$_2$(g) 和 H$_2$O(g)。通式： $$ \text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\,\text{CO}_2 + \frac{y}{2}\,\text{H}_2\text{O} $$
Incomplete combustion不完全燃烧 — limited O$_2$. Products include CO (carbon monoxide, toxic) and/or soot (C). SCH3U C3.2 requires students to explain the difference.— O$_2$ 有限。产物包括 CO（一氧化碳，有毒）和/或煤烟（C）。SCH3U C3.2 要求学生解释区别。
Combustion of other fuels其他燃料的燃烧
- Hydrogen: $2\,\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\,\text{H}_2\text{O(g)}$ — only product is water.氢气：$2\,\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\,\text{H}_2\text{O(g)}$ — 唯一产物是水。
- Sulfur: $\text{S(s)} + \text{O}_2\text{(g)} \to \text{SO}_2\text{(g)}$ — produces acid rain precursor.硫：$\text{S(s)} + \text{O}_2\text{(g)} \to \text{SO}_2\text{(g)}$ — 产生酸雨前体。
- Metals: $2\,\text{Mg(s)} + \text{O}_2\text{(g)} \to 2\,\text{MgO(s)}$ — metal oxide forms (also a synthesis).金属：$2\,\text{Mg(s)} + \text{O}_2\text{(g)} \to 2\,\text{MgO(s)}$ — 生成金属氧化物（也是化合反应）。

NGSS HS-PS1-2 Assessment Boundary explicitly includes combustion reactions. SCH3U C3.2 and AB Chemistry 30 Unit C discuss combustion of organic compounds. BC Chemistry 11 content "reactions: predicting products, reactants and energy changes (ΔH)" includes combustion.NGSS HS-PS1-2 评估边界明确包括燃烧反应。SCH3U C3.2 与 AB Chemistry 30 Unit C 讨论有机化合物的燃烧。BC Chemistry 11 内容"反应：预测产物、反应物与能量变化（ΔH）"包括燃烧。

Worked Example 5 · Balancing a combustion equation例题 5 · 配平燃烧方程式

Write the balanced equation for the complete combustion of propane (C$_3$H$_8$).写出丙烷（C$_3$H$_8$）完全燃烧的配平方程式。

Identify products of complete combustion:确定完全燃烧的产物： C$_3$H$_8$ + O$_2$ → CO$_2$ + H$_2$O.C$_3$H$_8$ + O$_2$ → CO$_2$ + H$_2$O。

Balance C:配平 C： 3 C in propane → 3 CO$_2$.丙烷中有 3 个 C → 3 CO$_2$。

Balance H:配平 H： 8 H in propane → 4 H$_2$O.丙烷中有 8 个 H → 4 H$_2$O。

Balance O:配平 O： Right side: $3 \times 2 + 4 \times 1 = 10$ O atoms → 5 O$_2$.右侧：$3 \times 2 + 4 \times 1 = 10$ 个 O 原子 → 5 O$_2$。

$$ \text{C}_3\text{H}_8\text{(g)} + 5\,\text{O}_2\text{(g)} \rightarrow 3\,\text{CO}_2\text{(g)} + 4\,\text{H}_2\text{O(g)} $$

Check:核验： Left: 3 C, 8 H, 10 O. Right: 3 C (in CO$_2$), 8 H (in H$_2$O), $6 + 4 = 10$ O. ✓左侧：3 C，8 H，10 O。右侧：3 C（在 CO$_2$ 中），8 H（在 H$_2$O 中），$6 + 4 = 10$ O。✓

What are the products of the complete combustion of octane (C$_8$H$_{18}$) in excess oxygen?辛烷（C$_8$H$_{18}$）在过量氧气中完全燃烧的产物是什么？

§5 · Q1

CO(g) and H$_2$O(g)CO(g) 和 H$_2$O(g)

CO$_2$(g) and H$_2$O(g)CO$_2$(g) 和 H$_2$O(g)

C(s) and H$_2$O(g)C(s) 和 H$_2$O(g)

CO$_2$(g) and H$_2$(g)CO$_2$(g) 和 H$_2$(g)

Complete combustion of any hydrocarbon in excess O$_2$ produces only CO$_2$ and H$_2$O. CO and soot (C) form only in incomplete combustion (limited O$_2$). H$_2$ is never a combustion product — it is a fuel.任何烃在过量 O$_2$ 中的完全燃烧只产生 CO$_2$ 和 H$_2$O。CO 和煤烟（C）仅在不完全燃烧（O$_2$ 有限）时形成。H$_2$ 从不是燃烧产物——它是燃料。

In complete combustion (excess O$_2$): all C → CO$_2$; all H → H$_2$O. Incomplete combustion (limited O$_2$) gives CO or soot. Hydrogen is a fuel, not a combustion product.在完全燃烧（过量 O$_2$）中：所有 C → CO$_2$；所有 H → H$_2$O。不完全燃烧（O$_2$ 有限）产生 CO 或煤烟。氢气是燃料，不是燃烧产物。

What coefficient is needed in front of O$_2$ to balance the complete combustion of ethanol (C$_2$H$_5$OH)?配平乙醇（C$_2$H$_5$OH）完全燃烧方程式时，O$_2$ 前需要的系数是多少？

§5 · Q2

C$_2$H$_5$OH + O$_2$ → 2 CO$_2$ + 3 H$_2$O. Balance C (2) and H (6 → 3 H$_2$O), then O: right side $= 4 + 3 = 7$ O; left has 1 O in ethanol, so O$_2$ must supply 6 O → coefficient 3. Full equation: C$_2$H$_5$OH(l) + 3 O$_2$(g) → 2 CO$_2$(g) + 3 H$_2$O(g).C$_2$H$_5$OH + O$_2$ → 2 CO$_2$ + 3 H$_2$O。配平 C（2）和 H（6 → 3 H$_2$O），然后配平 O：右侧 $= 4 + 3 = 7$ O；乙醇中有 1 O，故 O$_2$ 必须提供 6 O → 系数 3。完整方程式：C$_2$H$_5$OH(l) + 3 O$_2$(g) → 2 CO$_2$(g) + 3 H$_2$O(g)。

Balance in order: C first (2 CO$_2$), then H (6 H → 3 H$_2$O), then O last. Right side O: 4 + 3 = 7. Ethanol already has 1 O, so O$_2$ needs to supply 6 O = 3 O$_2$.按顺序配平：先 C（2 CO$_2$），再 H（6 H → 3 H$_2$O），最后 O。右侧 O：4 + 3 = 7。乙醇已有 1 O，故 O$_2$ 需提供 6 O = 3 O$_2$。

Going deeper — incomplete combustion and air quality深入 — 不完全燃烧与空气质量

Incomplete combustion occurs when the supply of oxygen is limited. Instead of the clean products CO$_2$ and H$_2$O, you get carbon monoxide (CO) and/or particulate carbon (soot). CO is a colourless, odourless, and highly toxic gas: it binds to haemoglobin about 200 times more strongly than O$_2$, blocking oxygen transport and causing suffocation at parts-per-million concentrations. Soot contributes to respiratory disease and climate forcing. SCH3U C3.2 explicitly requires students to "explain the difference between a complete combustion reaction and an incomplete combustion reaction." An example: a Bunsen burner with the air hole closed (yellow, luminous flame, soot) vs. open (blue, non-luminous, clean). Real-world implication: internal combustion engines require catalytic converters to oxidise CO to CO$_2$ before emission.当氧气供应有限时，发生不完全燃烧。不是产生干净的产物 CO$_2$ 和 H$_2$O，而是产生一氧化碳（CO）和/或颗粒碳（煤烟）。CO 是一种无色、无味、剧毒的气体：它与血红蛋白的结合强度约为 O$_2$ 的 200 倍，在百万分之几的浓度下就会阻断氧气运输并导致窒息。煤烟会导致呼吸系统疾病和气候强迫。SCH3U C3.2 明确要求学生"解释完全燃烧反应和不完全燃烧反应的区别"。示例：关闭气孔的本生灯（黄色、发光火焰、煤烟）与打开气孔（蓝色、非发光、清洁）的对比。现实意义：内燃机需要催化转化器在排放前将 CO 氧化为 CO$_2$。

Section 6 · Solubility rules and precipitation第 6 节 · 溶解度规则与沉淀

Solubility Rules and Precipitation Reactions溶解度规则与沉淀反应

Seven rules tell you whether an ionic compound dissolves or precipitates.七条规则告诉你离子化合物是溶解还是沉淀。

Ion / compound离子/化合物	Solubility溶解性	Exceptions例外
Group 1 (Li$^+$, Na$^+$, K$^+$, Rb$^+$, Cs$^+$) salts第 1 族（Li$^+$、Na$^+$、K$^+$、Rb$^+$、Cs$^+$）盐	Soluble可溶	None无
NH$_4^+$ saltsNH$_4^+$ 盐	Soluble可溶	None无
NO$_3^-$ saltsNO$_3^-$ 盐	Soluble可溶	None无
Cl$^-$, Br$^-$, I$^-$ saltsCl$^-$、Br$^-$、I$^-$ 盐	Soluble可溶	Ag$^+$, Pb$^{2+}$, Hg$_2^{2+}$ salts — insolubleAg$^+$、Pb$^{2+}$、Hg$_2^{2+}$ 盐 — 不溶
SO$_4^{2-}$ saltsSO$_4^{2-}$ 盐	Soluble可溶	Ba$^{2+}$, Pb$^{2+}$, Ca$^{2+}$, Sr$^{2+}$ — insolubleBa$^{2+}$、Pb$^{2+}$、Ca$^{2+}$、Sr$^{2+}$ — 不溶
OH$^-$ saltsOH$^-$ 盐	Insoluble不溶	Group 1, Ba$^{2+}$, Sr$^{2+}$ — soluble第 1 族、Ba$^{2+}$、Sr$^{2+}$ — 可溶
CO$_3^{2-}$, PO$_4^{3-}$, S$^{2-}$ saltsCO$_3^{2-}$、PO$_4^{3-}$、S$^{2-}$ 盐	Insoluble不溶	Group 1 and NH$_4^+$ — soluble第 1 族和 NH$_4^+$ — 可溶

A precipitation reaction is a double-replacement reaction in which one product is insoluble. Use the table to judge: if the predicted product is insoluble, write (s); if soluble, write (aq) and note no precipitate forms. SCH3U E3.4 and AB Chemistry 20 D GO1 both assess using a solubility table to predict precipitate formation.沉淀反应是一种双置换反应，其中一种产物不溶。使用上表判断：如果预测的产物不溶，写 (s)；如果可溶，写 (aq) 并注明无沉淀形成。SCH3U E3.4 与 AB Chemistry 20 D GO1 均考查使用溶解度表预测沉淀形成。

Worked Example 6 · Predicting whether a precipitate forms例题 6 · 预测是否生成沉淀

Does a precipitate form when aqueous lead(II) nitrate (Pb(NO$_3$)$_2$(aq)) and aqueous potassium iodide (KI(aq)) are mixed? Write the balanced molecular equation and identify the precipitate.将硝酸铅（II）（Pb(NO$_3$)$_2$(aq)）和碘化钾（KI(aq)）的水溶液混合时，是否生成沉淀？写出配平的分子方程式并确定沉淀。

Identify the ions:确定离子： Pb$^{2+}$, NO$_3^-$, K$^+$, I$^-$.Pb$^{2+}$, NO$_3^-$, K$^+$, I$^-$。

Swap partners:交换伙伴： Pb$^{2+}$ + I$^-$ → PbI$_2$; K$^+$ + NO$_3^-$ → KNO$_3$.Pb$^{2+}$ + I$^-$ → PbI$_2$；K$^+$ + NO$_3^-$ → KNO$_3$。

Check solubility:检查溶解性： PbI$_2$ — iodide of Pb$^{2+}$ — is an exception (insoluble, bright yellow precipitate). KNO$_3$ is soluble (Group 1 cation + nitrate).PbI$_2$ — Pb$^{2+}$ 的碘化物 — 是例外（不溶，亮黄色沉淀）。KNO$_3$ 可溶（第 1 族阳离子 + 硝酸根）。

Balanced molecular equation:配平分子方程式：

$$ \text{Pb(NO}_3)_2\text{(aq)} + 2\,\text{KI(aq)} \rightarrow \text{PbI}_2\text{(s)} + 2\,\text{KNO}_3\text{(aq)} $$

The bright yellow solid PbI$_2$ is the precipitate. ✓亮黄色固体 PbI$_2$ 是沉淀。✓

Aqueous solutions of sodium chloride (NaCl) and silver nitrate (AgNO$_3$) are mixed. What is the precipitate formed?氯化钠（NaCl）和硝酸银（AgNO$_3$）的水溶液混合。生成的沉淀是什么？

§6 · Q1

NaNO$_3$NaNO$_3$

No precipitate forms无沉淀形成

AgNO$_3$AgNO$_3$

AgClAgCl

The double-replacement products are AgCl and NaNO$_3$. AgCl is the insoluble exception for chlorides (Ag$^+$ with Cl$^-$ → white precipitate). NaNO$_3$ is soluble (Group 1 + nitrate). The balanced equation: NaCl(aq) + AgNO$_3$(aq) → AgCl(s) + NaNO$_3$(aq).双置换产物是 AgCl 和 NaNO$_3$。AgCl 是氯化物的不溶例外（Ag$^+$ 与 Cl$^-$ → 白色沉淀）。NaNO$_3$ 可溶（第 1 族 + 硝酸根）。配平方程式：NaCl(aq) + AgNO$_3$(aq) → AgCl(s) + NaNO$_3$(aq)。

Swap the cations: Ag$^+$ pairs with Cl$^-$ → AgCl; Na$^+$ pairs with NO$_3^-$ → NaNO$_3$. Chlorides are generally soluble, but Ag$^+$, Pb$^{2+}$, Hg$_2^{2+}$ chlorides are the key exceptions. AgCl is white and insoluble.交换阳离子：Ag$^+$ 与 Cl$^-$ 配对 → AgCl；Na$^+$ 与 NO$_3^-$ 配对 → NaNO$_3$。氯化物通常可溶，但 Ag$^+$、Pb$^{2+}$、Hg$_2^{2+}$ 氯化物是关键例外。AgCl 是白色不溶固体。

Which compound is insoluble in water according to the solubility rules?根据溶解度规则，以下哪种化合物不溶于水？

§6 · Q2

BaSO$_4$BaSO$_4$

Na$_2$SO$_4$Na$_2$SO$_4$

KOHKOH

NH$_4$ClNH$_4$Cl

BaSO$_4$ — sulfates are generally soluble, but Ba$^{2+}$ is a key exception (insoluble). Na$_2$SO$_4$ is soluble (Group 1). KOH is soluble (Group 1 hydroxide). NH$_4$Cl is soluble (NH$_4^+$ salts and chlorides of non-exception metals are soluble).BaSO$_4$ — 硫酸盐通常可溶，但 Ba$^{2+}$ 是关键例外（不溶）。Na$_2$SO$_4$ 可溶（第 1 族）。KOH 可溶（第 1 族氢氧化物）。NH$_4$Cl 可溶（NH$_4^+$ 盐和非例外金属的氯化物可溶）。

Check each: Na$_2$SO$_4$ (Group 1 → soluble), KOH (Group 1 hydroxide → soluble), NH$_4$Cl (NH$_4^+$ → soluble). BaSO$_4$ is the exception: Ba$^{2+}$ sulfate is insoluble (used in barium meals for medical imaging).逐一检查：Na$_2$SO$_4$（第 1 族 → 可溶），KOH（第 1 族氢氧化物 → 可溶），NH$_4$Cl（NH$_4^+$ → 可溶）。BaSO$_4$ 是例外：Ba$^{2+}$ 硫酸盐不溶（用于医学成像的钡餐）。

Section 7 · Net ionic equations第 7 节 · 净离子方程式

Net Ionic Equations净离子方程式

Three-step process: molecular → complete ionic → net ionic.三步流程：分子方程式 → 完整离子方程式 → 净离子方程式。

Step 1 — Molecular equation第 1 步——分子方程式 Write the balanced molecular equation with correct state symbols.写出带有正确状态符号的配平分子方程式。
Step 2 — Complete ionic equation第 2 步——完整离子方程式 Split all strong electrolytes in aqueous solution into their constituent ions. Keep molecular/covalent compounds (H$_2$O, gases), weak acids, and precipitates as undissociated formulas.将水溶液中所有强电解质拆分为其组成离子。将分子/共价化合物（H$_2$O、气体）、弱酸和沉淀保持为未解离的化学式。
Step 3 — Cancel spectator ions第 3 步——消去旁观离子 Spectator ions appear identically on both sides; cancel them. What remains is the net ionic equation — it shows only the species that actually participate in the chemical change.旁观离子在两侧完全相同；消去它们。剩余的就是净离子方程式——它只显示实际参与化学变化的物种。

AB Chemistry 20 D GO1 explicitly requires writing "balanced ionic and net ionic equations, including identification of spectator ions." SCH3U E2.5 requires "balanced net ionic equations to represent precipitation and neutralization reactions."AB Chemistry 20 D GO1 明确要求书写"配平的离子和净离子方程式，包括识别旁观离子"。SCH3U E2.5 要求"书写配平的净离子方程式以表示沉淀和中和反应"。

Worked Example 7 · Writing the net ionic equation for a precipitation reaction例题 7 · 书写沉淀反应的净离子方程式

Write the net ionic equation for: Na$_2$SO$_4$(aq) + BaCl$_2$(aq) → BaSO$_4$(s) + 2 NaCl(aq).书写下列反应的净离子方程式：Na$_2$SO$_4$(aq) + BaCl$_2$(aq) → BaSO$_4$(s) + 2 NaCl(aq)。

Step 1: Molecular equation is given.第 1 步：分子方程式已给出。

Step 2: Complete ionic equation — split all (aq) strong electrolytes into ions.第 2 步：完整离子方程式——将所有 (aq) 强电解质拆分为离子。

$$ 2\,\text{Na}^+\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} + \text{Ba}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)} \to \text{BaSO}_4\text{(s)} + 2\,\text{Na}^+\text{(aq)} + 2\,\text{Cl}^-\text{(aq)} $$

Step 3: Cancel spectator ions.第 3 步：消去旁观离子。 2 Na$^+$(aq) appears on both sides — cancel. 2 Cl$^-$(aq) appears on both sides — cancel.2 Na$^+$(aq) 在两侧出现——消去。2 Cl$^-$(aq) 在两侧出现——消去。

Net ionic equation:净离子方程式：

$$ \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} $$

This reveals that any source of Ba$^{2+}$ mixed with any source of SO$_4^{2-}$ gives the same reaction. ✓这揭示了任何 Ba$^{2+}$ 来源与任何 SO$_4^{2-}$ 来源混合都会产生同样的反应。✓

What are spectator ions in a net ionic equation?净离子方程式中的旁观离子是什么？

§7 · Q1

Ions that form the precipitate形成沉淀的离子

Ions produced only on the product side仅在产物侧产生的离子

Ions present in solution on both sides of the reaction that do not change在反应两侧的溶液中存在且不发生变化的离子

Ions that are consumed in the reaction在反应中被消耗的离子

Spectator ions appear identically (same formula, same phase, same coefficient) on both the reactant and product sides of the complete ionic equation. They do not participate in the chemical transformation — they are just "watching." Cancelling them gives the net ionic equation.旁观离子在完整离子方程式的反应物侧和产物侧以完全相同的形式（相同的化学式、相同的状态、相同的系数）出现。它们不参与化学转变——只是"旁观"。消去它们得到净离子方程式。

Precipitate-forming ions and consumed ions are NOT spectators — they change. Spectator ions are unchanged and appear identically on both sides. They are cancelled out in the step from complete ionic to net ionic equation.形成沉淀的离子和被消耗的离子不是旁观者——它们发生了变化。旁观离子不变，在两侧以完全相同的形式出现。从完整离子方程式到净离子方程式的步骤中，它们被消去。

The molecular equation is: 2 HCl(aq) + Ca(OH)$_2$(aq) → CaCl$_2$(aq) + 2 H$_2$O(l). What is the net ionic equation? (HCl and Ca(OH)$_2$ are strong electrolytes; CaCl$_2$ is soluble; H$_2$O is molecular.)分子方程式为：2 HCl(aq) + Ca(OH)$_2$(aq) → CaCl$_2$(aq) + 2 H$_2$O(l)。净离子方程式是什么？（HCl 和 Ca(OH)$_2$ 是强电解质；CaCl$_2$ 可溶；H$_2$O 是分子化合物。）

§7 · Q2

H$^+$(aq) + OH$^-$(aq) → H$_2$O(l)H$^+$(aq) + OH$^-$(aq) → H$_2$O(l)

2 HCl(aq) + Ca(OH)$_2$(aq) → CaCl$_2$(aq) + 2 H$_2$O(l)2 HCl(aq) + Ca(OH)$_2$(aq) → CaCl$_2$(aq) + 2 H$_2$O(l)

Ca$^{2+}$(aq) + 2 Cl$^-$(aq) → CaCl$_2$(aq)Ca$^{2+}$(aq) + 2 Cl$^-$(aq) → CaCl$_2$(aq)

2 H$^+$(aq) + Ca$^{2+}$(aq) + 2 OH$^-$(aq) → CaCl$_2$(aq) + 2 H$_2$O(l)2 H$^+$(aq) + Ca$^{2+}$(aq) + 2 OH$^-$(aq) → CaCl$_2$(aq) + 2 H$_2$O(l)

Complete ionic: 2 H$^+$(aq) + 2 Cl$^-$(aq) + Ca$^{2+}$(aq) + 2 OH$^-$(aq) → Ca$^{2+}$(aq) + 2 Cl$^-$(aq) + 2 H$_2$O(l). Spectator ions: Ca$^{2+}$ and 2 Cl$^-$ (appear on both sides). Cancel them. Net: H$^+$(aq) + OH$^-$(aq) → H$_2$O(l). This is the universal net ionic equation for any strong acid–strong base neutralization.完整离子方程式：2 H$^+$(aq) + 2 Cl$^-$(aq) + Ca$^{2+}$(aq) + 2 OH$^-$(aq) → Ca$^{2+}$(aq) + 2 Cl$^-$(aq) + 2 H$_2$O(l)。旁观离子：Ca$^{2+}$ 和 2 Cl$^-$（在两侧出现）。消去它们。净：H$^+$(aq) + OH$^-$(aq) → H$_2$O(l)。这是任何强酸–强碱中和的通用净离子方程式。

Split all (aq) strong electrolytes into ions. Then cancel species that appear identically on both sides (Ca$^{2+}$ and Cl$^-$ are spectators). What remains is H$^+$ + OH$^-$ → H$_2$O — the core of every strong-acid/strong-base neutralization.将所有 (aq) 强电解质拆分为离子。然后消去在两侧完全相同出现的物种（Ca$^{2+}$ 和 Cl$^-$ 是旁观者）。剩余的是 H$^+$ + OH$^-$ → H$_2$O——每个强酸/强碱中和的核心。

Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Balancing discipline (§2)配平纪律（§2）

Change only coefficients, never subscripts.只改变系数，不改变下标。 Changing H$_2$O to H$_4$O$_2$ changes the substance — that is not balancing, that is inventing a new compound. Use whole-number prefixes only.将 H$_2$O 改为 H$_4$O$_2$ 改变了物质——这不是配平，而是发明了一种新化合物。只使用整数前缀。
For combustion, save O$_2$ for last.对于燃烧，将 O$_2$ 留到最后配平。 Balance C first (from fuel to CO$_2$), then H (from fuel to H$_2$O), then count all O atoms on the right and determine how many O$_2$ molecules are needed.先配平 C（从燃料到 CO$_2$），再配平 H（从燃料到 H$_2$O），然后计算右侧所有 O 原子并确定需要多少 O$_2$ 分子。
Always verify the atom count.始终核验原子数。 After balancing, recount every element on both sides. A single coefficient error invalidates the whole equation.配平后，重新计算两侧每种元素的数量。一个系数错误会使整个方程式失效。

Reaction type identification (§3–§5)反应类型识别（§3–§5）

Count reactants and products first.首先计算反应物和产物的数量。 One reactant → decomposition. Two reactants, one product → synthesis. Two reactants, two products with element swap → replacement. Fuel + O$_2$ → oxides → combustion.一种反应物 → 分解。两种反应物，一种产物 → 化合。两种反应物，两种产物且有元素交换 → 置换。燃料 + O$_2$ → 氧化物 → 燃烧。
Single replacement: check the activity series before predicting.单置换：预测前检查活动序。 If the free metal (or halogen) is above the one in the compound in the activity series, the reaction is spontaneous. If below — no reaction.如果自由金属（或卤素）在活动序中位于化合物中金属的上方，则反应自发进行。如果在下方——无反应。

Solubility and precipitation (§6)溶解度与沉淀（§6）

Memorise the "always soluble" list.记住"总是可溶"列表。 Group 1 salts, NH$_4^+$ salts, NO$_3^-$ salts are always soluble — no exceptions. Halides (Cl$^-$, Br$^-$, I$^-$) are soluble except with Ag$^+$, Pb$^{2+}$, Hg$_2^{2+}$.第 1 族盐、NH$_4^+$ 盐、NO$_3^-$ 盐总是可溶——无例外。卤化物（Cl$^-$、Br$^-$、I$^-$）可溶，但与 Ag$^+$、Pb$^{2+}$、Hg$_2^{2+}$ 形成的盐不溶。
No precipitate = no reaction in double replacement.无沉淀 = 双置换中无反应。 If both products from swapping partners are soluble, write "no reaction" (NR). The reaction occurs only if at least one product is insoluble, a gas, or weakly ionized water.如果交换伙伴后两种产物都可溶，写"无反应"（NR）。只有当至少一种产物不溶、为气体或弱电离水时，反应才发生。

Net ionic equations (§7) — the three-step checklist净离子方程式（§7）——三步检查清单

Only split strong electrolytes in (aq).只拆分 (aq) 中的强电解质。 Strong acids (HCl, HNO$_3$, H$_2$SO$_4$, HBr, HI, HClO$_4$), strong bases (Group 1 hydroxides, Ba(OH)$_2$, Sr(OH)$_2$), and soluble ionic salts split into ions. Weak acids (CH$_3$COOH), precipitates (s), and water (l) stay as molecules.强酸（HCl、HNO$_3$、H$_2$SO$_4$、HBr、HI、HClO$_4$）、强碱（第 1 族氢氧化物、Ba(OH)$_2$、Sr(OH)$_2$）和可溶性离子盐拆分为离子。弱酸（CH$_3$COOH）、沉淀（s）和水（l）保持为分子。
Cancel symmetrically: same ion, same coefficient, same phase.对称消去：相同离子、相同系数、相同状态。 Do not cancel a 2 Na$^+$ on the left against a 1 Na$^+$ on the right — they must match exactly.不要将左侧的 2 Na$^+$ 与右侧的 1 Na$^+$ 相消——它们必须完全匹配。

Active Recall主动复习

Flashcards闪卡

5 signs of a chemical reaction?化学反应的 5 种证据？

Color change · gas produced · precipitate formed · energy change (heat/cold) · light emitted.颜色变化 · 产生气体 · 生成沉淀 · 能量变化（热/冷） · 发光。

Law of conservation of mass?质量守恒定律？

Atoms are rearranged, not created or destroyed. A balanced equation has equal atom counts on both sides.原子被重新排列，而不是被创造或消灭。配平方程式两侧每种原子的数量相等。

Synthesis?化合反应 (Synthesis)?

A + B → AB. Two or more reactants combine to form one product.
Example: $2\,\text{Mg} + \text{O}_2 \to 2\,\text{MgO}$A + B → AB。两种或更多反应物结合形成一种产物。
示例：$2\,\text{Mg} + \text{O}_2 \to 2\,\text{MgO}$

Decomposition?分解反应 (Decomposition)?

AB → A + B. One reactant breaks into two or more products. Often needs heat (Δ), light, or electricity.
Example: $2\,\text{H}_2\text{O}_2 \to 2\,\text{H}_2\text{O} + \text{O}_2$AB → A + B。一种反应物分解为两种或更多产物。通常需要热（Δ）、光或电。
示例：$2\,\text{H}_2\text{O}_2 \to 2\,\text{H}_2\text{O} + \text{O}_2$

Single replacement?单置换 (Single replacement)?

A + BC → AC + B. More active element displaces a less active one. Use activity series to decide if reaction occurs.
Example: $\text{Zn} + \text{CuSO}_4 \to \text{ZnSO}_4 + \text{Cu}$A + BC → AC + B。活动性较强的元素置换活动性较弱的元素。使用活动序判断反应是否发生。
示例：$\text{Zn} + \text{CuSO}_4 \to \text{ZnSO}_4 + \text{Cu}$

Double replacement?双置换 (Double replacement)?

AB + CD → AD + CB. Cations swap anion partners. Reaction occurs if a precipitate, gas, or H$_2$O forms.
Example: $\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \to \text{BaSO}_4\downarrow + 2\,\text{NaCl}$AB + CD → AD + CB。阳离子交换阴离子伙伴。如果产生沉淀、气体或 H$_2$O，则发生反应。
示例：$\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \to \text{BaSO}_4\downarrow + 2\,\text{NaCl}$

Complete combustion products?完全燃烧的产物？

Hydrocarbon + excess O$_2$ → CO$_2$(g) + H$_2$O(g).
Incomplete (limited O$_2$) → CO and/or C (soot) also form.烃 + 过量 O$_2$ → CO$_2$(g) + H$_2$O(g)。
不完全（O$_2$ 有限）→ 还会形成 CO 和/或 C（煤烟）。

Balancing combustion: order?配平燃烧：顺序？

1. Balance C → CO$_2$.
2. Balance H → H$_2$O.
3. Balance O$_2$ last.
Example: C$_3$H$_8$ + 5 O$_2$ → 3 CO$_2$ + 4 H$_2$O.1. 配平 C → CO$_2$。
2. 配平 H → H$_2$O。
3. 最后配平 O$_2$。
示例：C$_3$H$_8$ + 5 O$_2$ → 3 CO$_2$ + 4 H$_2$O。

Always-soluble ions?总是可溶的离子？

Group 1 cations · NH$_4^+$ · NO$_3^-$ — no exceptions.
Halides (Cl$^-$, Br$^-$, I$^-$) soluble except Ag$^+$, Pb$^{2+}$, Hg$_2^{2+}$.第 1 族阳离子 · NH$_4^+$ · NO$_3^-$ — 无例外。
卤化物（Cl$^-$、Br$^-$、I$^-$）可溶，除 Ag$^+$、Pb$^{2+}$、Hg$_2^{2+}$。

Usually-insoluble ions?通常不溶的离子？

OH$^-$, CO$_3^{2-}$, PO$_4^{3-}$, S$^{2-}$ — insoluble except with Group 1 and NH$_4^+$.
SO$_4^{2-}$ insoluble with Ba$^{2+}$, Pb$^{2+}$, Ca$^{2+}$, Sr$^{2+}$.OH$^-$、CO$_3^{2-}$、PO$_4^{3-}$、S$^{2-}$ — 不溶，除与第 1 族和 NH$_4^+$。
SO$_4^{2-}$ 与 Ba$^{2+}$、Pb$^{2+}$、Ca$^{2+}$、Sr$^{2+}$ 不溶。

Spectator ions?旁观离子？

Ions appearing identically (same formula, coefficient, phase) on both sides of the complete ionic equation. They do not participate — cancel them to get the net ionic equation.在完整离子方程式两侧以完全相同形式（相同化学式、系数、状态）出现的离子。它们不参与——消去它们得到净离子方程式。

Net ionic eq. for any strong acid + strong base?任何强酸 + 强碱的净离子方程式？

$$\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \to \text{H}_2\text{O(l)}$$ Always the same regardless of which acid or base, because the cation and anion of the salt are spectators.无论使用哪种酸或碱，结果始终相同，因为盐的阳离子和阴离子是旁观者。

Net ionic eq. for precipitation of BaSO$_4$?BaSO$_4$ 沉淀的净离子方程式？

$$\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \to \text{BaSO}_4\text{(s)}$$ All other ions (e.g. Na$^+$, Cl$^-$) are spectators and cancel.所有其他离子（如 Na$^+$、Cl$^-$）是旁观者并被消去。

Balanced equation: key rule配平化学方程式 — 关键规则？

Change coefficients only — never subscripts. Verify by counting each element on both sides. Lowest whole-number ratios.只改变系数——不改变下标。通过计算两侧每种元素来核验。使用最低整数比。

Mixed Practice综合练习

Practice Quiz综合测验

A student heats copper(II) carbonate powder (CuCO$_3$) and observes a colour change from green to black, plus gas evolution. What type of reaction is this? 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.1学生加热碳酸铜粉末（CuCO$_3$），观察到颜色从绿色变为黑色，并有气体逸出。这是哪种类型的反应？🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.1

Synthesis化合反应

Combustion燃烧

Decomposition分解反应

Single replacement单置换

CuCO$_3$(s) → CuO(s) + CO$_2$(g). One reactant (CuCO$_3$) breaks into two products (CuO and CO$_2$) — the AB → A + B pattern. The black product is CuO; the gas is CO$_2$. This is thermal decomposition.CuCO$_3$(s) → CuO(s) + CO$_2$(g)。一种反应物（CuCO$_3$）分解为两种产物（CuO 和 CO$_2$）——AB → A + B 的模式。黑色产物是 CuO；气体是 CO$_2$。这是热分解。

Decomposition: one reactant → multiple products (no additional reactants required). The evidence — heat needed, gas and solid formed from one solid — all point to decomposition. Synthesis would combine reactants; combustion needs O$_2$.分解反应：一种反应物 → 多种产物（不需要额外的反应物）。证据——需要热量，从一种固体形成气体和固体——都指向分解反应。化合反应会结合反应物；燃烧需要 O$_2$。

Iron (Fe) is placed in a solution of copper(II) nitrate (Cu(NO$_3$)$_2$). The activity series (most to least active) includes: Mg, Al, Zn, Fe, Ni, Cu. What is observed? 🇨🇦 SCH3U C2.5 / AB Chem 20 D GO1将铁（Fe）放入硝酸铜（II）溶液（Cu(NO$_3$)$_2$）中。活动序（从最活泼到最不活泼）包括：Mg、Al、Zn、Fe、Ni、Cu。会观察到什么？🇨🇦 SCH3U C2.5 / AB Chem 20 D GO1

No reaction because iron is less reactive than copper无反应，因为铁比铜活动性低

Copper dissolves and iron nitrate forms铜溶解，形成硝酸铁

A blue precipitate forms形成蓝色沉淀

A red-brown solid deposits on the iron and the blue colour fades铁上析出红棕色固体，蓝色褪去

Fe is above Cu in the activity series, so Fe displaces Cu$^{2+}$: Fe(s) + Cu(NO$_3$)$_2$(aq) → Fe(NO$_3$)$_2$(aq) + Cu(s). Cu deposits as a red-brown solid; the blue Cu$^{2+}$ colour fades as colourless Fe$^{2+}$ forms. This is single replacement.Fe 在活动序中高于 Cu，故 Fe 置换 Cu$^{2+}$：Fe(s) + Cu(NO$_3$)$_2$(aq) → Fe(NO$_3$)$_2$(aq) + Cu(s)。Cu 以红棕色固体析出；蓝色 Cu$^{2+}$ 褪去，因为无色 Fe$^{2+}$ 形成。这是单置换反应。

Fe is above Cu in the activity series — Fe is more reactive. It displaces Cu from solution, depositing copper metal. A blue precipitate would not form; the solution loses blue colour as Cu$^{2+}$ is consumed.Fe 在活动序中高于 Cu——Fe 更活泼。它从溶液中置换出铜，沉积铜金属。不会形成蓝色沉淀；随着 Cu$^{2+}$ 被消耗，溶液失去蓝色。

What are the products of complete combustion of butane (C$_4$H$_{10}$)? 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.2丁烷（C$_4$H$_{10}$）完全燃烧的产物是什么？🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.2

CO$_2$(g) and H$_2$O(g)CO$_2$(g) 和 H$_2$O(g)

CO(g) and H$_2$O(g)CO(g) 和 H$_2$O(g)

C(s) and H$_2$(g)C(s) 和 H$_2$(g)

CO$_2$(g) and H$_2$(g)CO$_2$(g) 和 H$_2$(g)

Complete combustion of any hydrocarbon in excess O$_2$ always yields only CO$_2$ and H$_2$O. CO and soot form only when O$_2$ is limited (incomplete combustion). Balanced: 2 C$_4$H$_{10}$(g) + 13 O$_2$(g) → 8 CO$_2$(g) + 10 H$_2$O(g).任何烃在过量 O$_2$ 中的完全燃烧总是只产生 CO$_2$ 和 H$_2$O。CO 和煤烟只有在 O$_2$ 有限时（不完全燃烧）才会形成。配平：2 C$_4$H$_{10}$(g) + 13 O$_2$(g) → 8 CO$_2$(g) + 10 H$_2$O(g)。

In complete combustion (excess O$_2$): all C atoms → CO$_2$; all H atoms → H$_2$O. The defining feature is excess oxygen. H$_2$ and soot (C) appear only in thermal decomposition, not in combustion.在完全燃烧（过量 O$_2$）中：所有 C 原子 → CO$_2$；所有 H 原子 → H$_2$O。决定性特征是过量氧气。H$_2$ 和煤烟（C）只出现在热分解中，不在燃烧中。

Which pair of solutions, when mixed, will produce a precipitate? 🇨🇦 SCH3U E3.4 / AB Chem 20 D GO1以下哪对溶液混合后会产生沉淀？🇨🇦 SCH3U E3.4 / AB Chem 20 D GO1

NaCl(aq) + KNO$_3$(aq)NaCl(aq) + KNO$_3$(aq)

Pb(NO$_3$)$_2$(aq) + Na$_2$SO$_4$(aq)Pb(NO$_3$)$_2$(aq) + Na$_2$SO$_4$(aq)

KCl(aq) + NaNO$_3$(aq)KCl(aq) + NaNO$_3$(aq)

Na$_2$CO$_3$(aq) + K$_2$SO$_4$(aq)Na$_2$CO$_3$(aq) + K$_2$SO$_4$(aq)

Pb(NO$_3$)$_2$ + Na$_2$SO$_4$ → PbSO$_4$(s) + 2 NaNO$_3$(aq). PbSO$_4$ is insoluble (Pb$^{2+}$ is one of the exceptions for sulfate). Options A, B, C give all-soluble products (NaNO$_3$, KNO$_3$, NaCl all soluble). Option D: K$_2$CO$_3$ and Na$_2$SO$_4$ swap to give Na$_2$CO$_3$ and K$_2$SO$_4$ — all Group 1, all soluble.Pb(NO$_3$)$_2$ + Na$_2$SO$_4$ → PbSO$_4$(s) + 2 NaNO$_3$(aq)。PbSO$_4$ 不溶（Pb$^{2+}$ 是硫酸盐的例外之一）。选项 A、B、C 给出全部可溶的产物（NaNO$_3$、KNO$_3$、NaCl 均可溶）。选项 D：K$_2$CO$_3$ 和 Na$_2$SO$_4$ 交换给出 Na$_2$CO$_3$ 和 K$_2$SO$_4$——全部是第 1 族，全部可溶。

Swap the ion partners and test each product for solubility. Only Pb$^{2+}$ + SO$_4^{2-}$ → PbSO$_4$(s) produces an insoluble compound. All other combinations give Group 1 salts or nitrates — both always soluble.交换离子伙伴并测试每种产物的溶解性。只有 Pb$^{2+}$ + SO$_4^{2-}$ → PbSO$_4$(s) 产生不溶化合物。所有其他组合给出第 1 族盐或硝酸盐——两者总是可溶。

The molecular equation is: Na$_2$CO$_3$(aq) + 2 HCl(aq) → 2 NaCl(aq) + H$_2$O(l) + CO$_2$(g). What is the correct net ionic equation? (HCl, Na$_2$CO$_3$, and NaCl are strong electrolytes.) 🇨🇦 SCH3U E2.5 / AB Chem 20 D GO1分子方程式为：Na$_2$CO$_3$(aq) + 2 HCl(aq) → 2 NaCl(aq) + H$_2$O(l) + CO$_2$(g)。正确的净离子方程式是什么？（HCl、Na$_2$CO$_3$ 和 NaCl 是强电解质。）🇨🇦 SCH3U E2.5 / AB Chem 20 D GO1

Na$_2$CO$_3$(aq) + 2 HCl(aq) → 2 NaCl(aq) + H$_2$O(l) + CO$_2$(g)Na$_2$CO$_3$(aq) + 2 HCl(aq) → 2 NaCl(aq) + H$_2$O(l) + CO$_2$(g)

Na$^+$(aq) + Cl$^-$(aq) → NaCl(aq)Na$^+$(aq) + Cl$^-$(aq) → NaCl(aq)

CO$_3^{2-}$(aq) + 2 H$^+$(aq) → H$_2$O(l) + CO$_2$(g)CO$_3^{2-}$(aq) + 2 H$^+$(aq) → H$_2$O(l) + CO$_2$(g)

2 Na$^+$(aq) + CO$_3^{2-}$(aq) + 2 H$^+$(aq) + 2 Cl$^-$(aq) → products2 Na$^+$(aq) + CO$_3^{2-}$(aq) + 2 H$^+$(aq) + 2 Cl$^-$(aq) → 产物

Complete ionic: 2 Na$^+$ + CO$_3^{2-}$ + 2 H$^+$ + 2 Cl$^-$ → 2 Na$^+$ + 2 Cl$^-$ + H$_2$O + CO$_2$. Spectators: 2 Na$^+$ and 2 Cl$^-$ (cancel). Net: CO$_3^{2-}$(aq) + 2 H$^+$(aq) → H$_2$O(l) + CO$_2$(g). H$_2$O and CO$_2$ are molecular/gas — not split.完整离子方程式：2 Na$^+$ + CO$_3^{2-}$ + 2 H$^+$ + 2 Cl$^-$ → 2 Na$^+$ + 2 Cl$^-$ + H$_2$O + CO$_2$。旁观者：2 Na$^+$ 和 2 Cl$^-$（消去）。净：CO$_3^{2-}$(aq) + 2 H$^+$(aq) → H$_2$O(l) + CO$_2$(g)。H$_2$O 和 CO$_2$ 是分子/气体——不拆分。

Split strong electrolytes in (aq): Na$_2$CO$_3$ → 2 Na$^+$ + CO$_3^{2-}$; HCl → H$^+$ + Cl$^-$; NaCl → Na$^+$ + Cl$^-$. H$_2$O and CO$_2$ stay as molecules. Then cancel Na$^+$ and Cl$^-$ from both sides.拆分 (aq) 中的强电解质：Na$_2$CO$_3$ → 2 Na$^+$ + CO$_3^{2-}$；HCl → H$^+$ + Cl$^-$；NaCl → Na$^+$ + Cl$^-$。H$_2$O 和 CO$_2$ 保持为分子。然后从两侧消去 Na$^+$ 和 Cl$^-$。

Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

0 / 11 mastered已掌握 0 / 11

✓List five observable pieces of evidence that a chemical reaction has occurred and explain why each indicates a new substance has formed. 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C2.1列出五种化学反应发生的可观测证据，并解释为什么每种证据表明新物质已形成。🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C2.1
✓Balance any chemical equation using only coefficients, verify atom counts on both sides, and write correct state symbols (s, l, g, aq). 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U C2.2 / BC Sci 10只使用系数配平任何化学方程式，核验两侧的原子数，并写出正确的状态符号（s、l、g、aq）。🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U C2.2 / BC Sci 10
✓Identify and write balanced equations for synthesis and decomposition reactions, predicting the single product or the products from the reactant. 🇨🇦 SCH3U C3.1 / AB Chem 20 D GO1识别并书写化合反应和分解反应的配平方程式，从反应物预测唯一产物或各产物。🇨🇦 SCH3U C3.1 / AB Chem 20 D GO1
✓Use the metal activity series to determine whether a single replacement reaction is spontaneous, and write the balanced equation for the products. 🇨🇦 SCH3U C2.5 / AB Chem 20 D GO1使用金属活动序判断单置换反应是否自发进行，并写出产物的配平方程式。🇨🇦 SCH3U C2.5 / AB Chem 20 D GO1
✓Predict and write the products of double-replacement reactions, explaining why the reaction occurs (precipitate, gas, or water formed). 🇨🇦 SCH3U C2.6 / AB Chem 20 D GO1预测并写出双置换反应的产物，解释为什么发生反应（生成沉淀、气体或水）。🇨🇦 SCH3U C2.6 / AB Chem 20 D GO1
✓Balance the combustion of any hydrocarbon or alcohol by balancing C, then H, then O$_2$ last; distinguish complete (CO$_2$ + H$_2$O) from incomplete combustion (CO or C). 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.2 / BC Chem 11通过先配平 C，再配平 H，最后配平 O$_2$ 来配平任何烃或醇的燃烧（燃烧）方程式；区分完全燃烧（CO$_2$ + H$_2$O）和不完全燃烧（CO 或 C）。🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.2 / BC Chem 11
✓Apply solubility rules to predict whether a precipitate forms when two aqueous solutions are mixed; if so, identify the precipitate and write the molecular equation. 🇨🇦 SCH3U E3.4 / AB Chem 20 D GO1应用溶解度规则预测两种水溶液混合时是否生成沉淀；如果是，识别沉淀并写出分子方程式。🇨🇦 SCH3U E3.4 / AB Chem 20 D GO1
✓Convert a molecular equation to a complete ionic equation by splitting all strong electrolytes in (aq) into ions; keep precipitates, gases, and water as molecules. 🇨🇦 SCH3U E2.5 / AB Chem 20 D GO1将分子方程式转换为完整离子方程式，将所有 (aq) 强电解质拆分为离子；将沉淀、气体和水保持为分子。🇨🇦 SCH3U E2.5 / AB Chem 20 D GO1
✓Write the net ionic equation by identifying and cancelling spectator ions; verify it is balanced in both atoms and charge. 🇨🇦 SCH3U E2.5 / AB Chem 20 D GO1通过识别和消去旁观离子写出净离子方程式；核验原子数和电荷均已配平。🇨🇦 SCH3U E2.5 / AB Chem 20 D GO1
✓State the net ionic equation for strong acid + strong base neutralization: H$^+$(aq) + OH$^-$(aq) → H$_2$O(l); explain why it is the same regardless of which acid and base are used. 🇨🇦 SCH3U E2.5 / AB Chem 20 D GO2陈述强酸 + 强碱中和的净离子方程式：H$^+$(aq) + OH$^-$(aq) → H$_2$O(l)；解释为什么无论使用哪种酸和碱结果都相同。🇨🇦 SCH3U E2.5 / AB Chem 20 D GO2
✓Classify a given balanced equation as synthesis, decomposition, single replacement, double replacement, or combustion, and justify the classification with the general form it matches. 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.1 / BC Chem 11将给定的配平方程式分类为化合、分解、单置换、双置换或燃烧反应，并用其匹配的通式为分类提供理由。🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C3.1 / BC Chem 11

Next Steps后续单元

What This Feeds Into本单元的去向

Chemical reactions and equations is the procedural core that every subsequent chemistry topic assumes. The classification and balancing skills here feed directly into stoichiometry (mole ratios in balanced equations), acid-base chemistry (neutralization as a double replacement), solutions and solubility (precipitation and net ionic equations), kinetics (monitoring reactions via evidence of change), and equilibrium (reversible reactions). The cross-references below point at the college-credit feeder and the next High School Chemistry unit.化学反应与方程式是每个后续化学主题所需的程序性核心。这里的分类和配平技能直接供给化学计量学（配平方程式中的摩尔比）、酸碱化学（中和作为双置换）、溶液与溶解度（沉淀和净离子方程式）、动力学（通过变化证据监测反应）和平衡（可逆反应）。以下链接指向大学学分衔接课程和下一个高中化学单元。

Within High School Chemistry.在 HS Chemistry 内部。

The next unit, The Mole and Stoichiometry (Unit 5), applies balanced equations quantitatively — mole ratios, limiting reagents, and percent yield all require a correctly balanced equation as the starting point. Solutions and Solubility (Unit 8) extends §6 precipitation to concentration calculations and quantitative solubility. Acids, Bases and pH (Unit 9) builds the neutralization reaction from the net ionic equation H$^+$ + OH$^-$ → H$_2$O. Every quantitative unit in the course begins from the balanced equation you write here.下一个单元《摩尔与化学计量学》（第 5 单元）定量应用配平方程式——摩尔比、限量试剂和百分产率都需要正确配平的方程式作为起点。《溶液与溶解度》（第 8 单元）将 §6 的沉淀扩展到浓度计算和定量溶解度。《酸、碱与 pH》（第 9 单元）从净离子方程式 H$^+$ + OH$^-$ → H$_2$O 建立中和反应。课程中每个定量单元都从你在这里写的配平方程式开始。

College-credit feeder and cross-subject links.大学学分衔接与跨学科链接。

IB Chemistry HL · Reactivity 2: How Much, How Fast, How Far? (the college-credit feeder — reaction types, stoichiometry, reaction rates, and equilibrium all build on the reaction-classification and net-ionic-equation skills here)IB Chemistry HL · Reactivity 2：多少？多快？多远？（大学学分衔接——反应类型、化学计量学、反应速率和平衡都建立在这里的反应分类和净离子方程式技能之上）

If you are aiming for IB Chemistry HL or AP Chemistry, the ability to classify reactions, write and balance equations, and produce net ionic equations is assumed from the first class. IB Reactivity 2 extends this with reaction rates (collision theory), equilibrium constants (Keq), and acid-base equilibria. AP Chemistry Unit 4 (Chemical Reactions) requires predict-then-balance mastery for all five reaction types plus redox. The net ionic equation skill you build here is also the entry point for electrochemistry (Unit 13), where half-reactions are a specialised form of net ionic equations.如果你目标是 IB Chemistry HL 或 AP Chemistry，分类反应、书写和配平方程式以及写出净离子方程式的能力从第一堂课就被默认掌握。IB Reactivity 2 将此延伸至反应速率（碰撞理论）、平衡常数（Keq）和酸碱平衡。AP Chemistry Unit 4（化学反应）要求对所有五种反应类型加上氧化还原的预测-配平能力。你在这里建立的净离子方程式技能也是电化学（第 13 单元）的切入点，其中半反应是净离子方程式的一种特殊形式。