PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 24 marksAP 风格选择题 + 安/卑省考短答 · 共 24 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter and show sufficient working in the margin. For short-answer items, state units in every answer. Use $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$ throughout. No calculator on Q1-Q2; calculator permitted on Q3-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在旁边写出足以让阅卷人核对的过程。短答题每道都要写出单位。全卷取 $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。
Q1EASY易🇺🇸 US美AP-style MCQAP 风格选择题§1 Mole and Avogadro's number摩尔与阿伏伽德罗常数[3 marks][3 分]
How many molecules are in $2.00\ \mathrm{mol}$ of carbon dioxide ($\mathrm{CO_2}$)? ($N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$)$2.00\ \mathrm{mol}$ 二氧化碳（$\mathrm{CO_2}$）中含有多少个分子？（$N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$）
(A) $6.022 \times 10^{23}$
(B) $1.204 \times 10^{24}$
(C) $3.011 \times 10^{23}$
(D) $2.409 \times 10^{24}$
Q2EASY易🇺🇸 US美AP-style MCQAP 风格选择题§2 Molar mass and mole-mass conversion摩尔质量与摩尔-质量换算[3 marks][3 分]
How many moles are in $36.0\ \mathrm{g}$ of water ($\mathrm{H_2O}$, $M = 18.02\ \mathrm{g\,mol^{-1}}$)?$36.0\ \mathrm{g}$ 水（$\mathrm{H_2O}$，$M = 18.02\ \mathrm{g\,mol^{-1}}$）中含有多少摩尔？
Sodium chloride (NaCl) has a molar mass of $58.44\ \mathrm{g\,mol^{-1}}$.氯化钠（NaCl）的摩尔质量为 $58.44\ \mathrm{g\,mol^{-1}}$。
(a)Calculate the mass of $0.250\ \mathrm{mol}$ of NaCl, with units.计算 $0.250\ \mathrm{mol}$ NaCl 的质量，并写出单位。[2]
(b)Calculate the number of sodium ions in $0.250\ \mathrm{mol}$ of NaCl.计算 $0.250\ \mathrm{mol}$ NaCl 中钠离子的数目。[1]
(c)State one assumption made in part (b).写出 (b) 小问所依赖的一个假设。[1]
Q4MEDIUM中🇺🇸 US美AP-style MCQAP 风格选择题§4 Balancing chemical equations配平化学方程式[3 marks][3 分]
Which of the following is the correctly balanced equation for the reaction of aluminium with oxygen to form aluminium oxide ($\mathrm{Al_2O_3}$)?下列哪项是铝与氧气反应生成氧化铝（$\mathrm{Al_2O_3}$）的正确配平方程式？
Ammonium nitrate ($\mathrm{NH_4NO_3}$) is a common nitrogen-containing fertiliser. Its molar mass is $80.04\ \mathrm{g\,mol^{-1}}$.硝酸铵（$\mathrm{NH_4NO_3}$）是一种常见的含氮肥料，其摩尔质量为 $80.04\ \mathrm{g\,mol^{-1}}$。
(a)Determine the percent by mass of nitrogen in $\mathrm{NH_4NO_3}$. Show your calculation.计算 $\mathrm{NH_4NO_3}$ 中氮的质量分数，写出计算过程。[3]
(b)Calculate the mass of nitrogen in a $500.\ \mathrm{g}$ bag of pure $\mathrm{NH_4NO_3}$.计算 $500.\ \mathrm{g}$ 纯 $\mathrm{NH_4NO_3}$ 中氮的质量。[2]
(c)A compound is found to contain $40.0\%$ C, $6.67\%$ H, and $53.3\%$ O by mass. Determine its empirical formula. Show all working.已知某化合物按质量计含 $40.0\%$ C、$6.67\%$ H 和 $53.3\%$ O。求其实验式，写出全部过程。[4]
(d)Explain why two different compounds can share the same empirical formula.解释为何两种不同化合物可以有相同的实验式。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 33 marksAP 衔接简答题 + 荣誉级 · 共 33 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of your reasoning. Identify the mole ratio or conversion factor used before substituting values. State units in every final answer. Calculator permitted on Q6-Q9.每一步推理都要写出。在代入数值前先注明所用摩尔比或换算因子。每个最终答案都要写单位。Q6-Q9 可用计算器。
The Haber process synthesises ammonia by the reaction $\mathrm{N_2(g) + 3\,H_2(g) \to 2\,NH_3(g)}$.哈伯法通过反应 $\mathrm{N_2(g) + 3\,H_2(g) \to 2\,NH_3(g)}$ 合成氨气。
(a)State the mole ratio of $\mathrm{H_2}$ to $\mathrm{NH_3}$ in this reaction.写出该反应中 $\mathrm{H_2}$ 与 $\mathrm{NH_3}$ 的摩尔比。[1]
(b)Calculate the number of moles of $\mathrm{H_2}$ in $5.40\ \mathrm{g}$ of $\mathrm{H_2}$ ($M = 2.016\ \mathrm{g\,mol^{-1}}$).计算 $5.40\ \mathrm{g}$ $\mathrm{H_2}$（$M = 2.016\ \mathrm{g\,mol^{-1}}$）中的摩尔数。[2]
(c)Calculate the mass of $\mathrm{NH_3}$ produced when $5.40\ \mathrm{g}$ of $\mathrm{H_2}$ reacts completely with excess $\mathrm{N_2}$. ($M_{\mathrm{NH_3}} = 17.03\ \mathrm{g\,mol^{-1}}$)计算 $5.40\ \mathrm{g}$ $\mathrm{H_2}$ 与过量 $\mathrm{N_2}$ 完全反应时生成 $\mathrm{NH_3}$ 的质量。（$M_{\mathrm{NH_3}} = 17.03\ \mathrm{g\,mol^{-1}}$）[3]
(d)Calculate the mass of $\mathrm{N_2}$ consumed in part (c). ($M_{\mathrm{N_2}} = 28.02\ \mathrm{g\,mol^{-1}}$)计算 (c) 小问中消耗的 $\mathrm{N_2}$ 的质量。（$M_{\mathrm{N_2}} = 28.02\ \mathrm{g\,mol^{-1}}$）[2]
Q7MEDIUM中🇨🇦 ON安ON Provincial-style安大略省考风格§3 Empirical and molecular formula实验式与分子式[8 marks][8 分]
A compound is analysed and found to contain $40.0\%$ C, $6.67\%$ H, and $53.3\%$ O by mass. A separate experiment determines that its molar mass is approximately $60.1\ \mathrm{g\,mol^{-1}}$.分析一种化合物，发现其按质量计含 $40.0\%$ C、$6.67\%$ H 和 $53.3\%$ O。另一实验测定其摩尔质量约为 $60.1\ \mathrm{g\,mol^{-1}}$。
(a)Assume a $100.0\ \mathrm{g}$ sample. Convert each element mass to moles. Show full working.假设取 $100.0\ \mathrm{g}$ 样品，将每种元素的质量转换为摩尔数，写出完整过程。[3]
(b)Determine the empirical formula by finding the simplest whole-number mole ratio.通过求最简整数摩尔比确定实验式。[2]
(c)Use the molar mass to determine the molecular formula.利用摩尔质量确定分子式。[2]
(d)Identify one possible compound with this molecular formula.写出一种具有该分子式的可能化合物。[1]
Q8HARD难🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§6 Limiting and excess reactants限量反应物与过量反应物[9 marks][9 分]
Aluminium reacts with hydrochloric acid according to the equation $\mathrm{2\,Al(s) + 6\,HCl(aq) \to 2\,AlCl_3(aq) + 3\,H_2(g)}$. A student combines $5.40\ \mathrm{g}$ of Al with $14.6\ \mathrm{g}$ of HCl. ($M_{\mathrm{Al}} = 26.98$, $M_{\mathrm{HCl}} = 36.46\ \mathrm{g\,mol^{-1}}$)铝与盐酸按方程式 $\mathrm{2\,Al(s) + 6\,HCl(aq) \to 2\,AlCl_3(aq) + 3\,H_2(g)}$ 发生反应。一名学生将 $5.40\ \mathrm{g}$ Al 与 $14.6\ \mathrm{g}$ HCl 混合。（$M_{\mathrm{Al}} = 26.98$，$M_{\mathrm{HCl}} = 36.46\ \mathrm{g\,mol^{-1}}$）
(a)Calculate the number of moles of each reactant.计算每种反应物的摩尔数。[2]
(b)Identify the limiting reactant. Show the comparison that justifies your choice.确定限量反应物。写出支持你选择的比较计算。[3]
(c)Calculate the theoretical mass of $\mathrm{H_2}$ produced. ($M_{\mathrm{H_2}} = 2.016\ \mathrm{g\,mol^{-1}}$)计算生成 $\mathrm{H_2}$ 的理论质量。（$M_{\mathrm{H_2}} = 2.016\ \mathrm{g\,mol^{-1}}$）[2]
(d)Calculate the mass of the excess reactant remaining after the reaction.计算反应结束后剩余的过量反应物的质量。[2]
Q9HARD难Honors荣誉级🇺🇸 US美AP-feeder FRQAP 衔接简答题§3 Molecular formula from empirical formula and molar mass由实验式和摩尔质量推导分子式[8 marks][8 分]
A combustion analysis of $0.460\ \mathrm{g}$ of an unknown organic compound yields $0.879\ \mathrm{g}$ of $\mathrm{CO_2}$ and $0.540\ \mathrm{g}$ of $\mathrm{H_2O}$. A mass spectrometer measures its molar mass as $46.1\ \mathrm{g\,mol^{-1}}$. ($M_{\mathrm{CO_2}} = 44.01$, $M_{\mathrm{H_2O}} = 18.02\ \mathrm{g\,mol^{-1}}$)对 $0.460\ \mathrm{g}$ 未知有机化合物进行燃烧分析，得到 $0.879\ \mathrm{g}$ $\mathrm{CO_2}$ 和 $0.540\ \mathrm{g}$ $\mathrm{H_2O}$。质谱仪测得其摩尔质量为 $46.1\ \mathrm{g\,mol^{-1}}$。（$M_{\mathrm{CO_2}} = 44.01$，$M_{\mathrm{H_2O}} = 18.02\ \mathrm{g\,mol^{-1}}$）
(a)Calculate the moles of C and the moles of H from the combustion products.由燃烧产物计算 C 和 H 的摩尔数。[2]
(b)Determine the mass of oxygen in the original sample by subtraction. Then calculate the moles of O.用差减法确定原样品中氧的质量，再计算 O 的摩尔数。[2]
(c)Find the empirical formula and then the molecular formula. Show the multiplier calculation.求实验式，再求分子式，写出倍数计算过程。[3]
(d)Name one compound consistent with this molecular formula.写出一种与该分子式相符的化合物名称。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 25 marks阿省毕业考 + 通用题型 · 共 25 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all symbols (with units) at the start of each question. State the mole ratio or conversion step used before substituting. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义所有符号（含单位）。代入数值前先写出所用摩尔比或换算步骤。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
Calcium carbonate decomposes on heating: $\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}$. A student heats $25.0\ \mathrm{g}$ of $\mathrm{CaCO_3}$ and collects $11.2\ \mathrm{g}$ of $\mathrm{CaO}$. ($M_{\mathrm{CaCO_3}} = 100.09$, $M_{\mathrm{CaO}} = 56.08\ \mathrm{g\,mol^{-1}}$)碳酸钙加热分解：$\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}$。一名学生加热 $25.0\ \mathrm{g}$ $\mathrm{CaCO_3}$ 后收集到 $11.2\ \mathrm{g}$ $\mathrm{CaO}$。（$M_{\mathrm{CaCO_3}} = 100.09$，$M_{\mathrm{CaO}} = 56.08\ \mathrm{g\,mol^{-1}}$）
(a)Calculate the moles of $\mathrm{CaCO_3}$ used.计算使用的 $\mathrm{CaCO_3}$ 的摩尔数。[1]
(b)Calculate the theoretical yield of $\mathrm{CaO}$ in grams.计算 $\mathrm{CaO}$ 的理论产量（以克为单位）。[3]
(c)Calculate the percent yield of $\mathrm{CaO}$.计算 $\mathrm{CaO}$ 的产率。[2]
(d)Suggest one reason the percent yield is less than $100\%$.提出一个产率低于 $100\%$ 的可能原因。[2]
Q11MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§5 + §6 + §7 Multi-step stoichiometry with limiting reagent and percent yield含限量试剂与产率的多步化学计量[9 marks][9 分]
Iron is extracted from iron(III) oxide by reduction with carbon monoxide: $\mathrm{Fe_2O_3(s) + 3\,CO(g) \to 2\,Fe(s) + 3\,CO_2(g)}$. A smelter charges $160.0\ \mathrm{g}$ of $\mathrm{Fe_2O_3}$ and $84.0\ \mathrm{g}$ of CO. The actual yield of iron is $97.0\ \mathrm{g}$. ($M_{\mathrm{Fe_2O_3}} = 159.69$, $M_{\mathrm{CO}} = 28.01$, $M_{\mathrm{Fe}} = 55.85\ \mathrm{g\,mol^{-1}}$)铁由一氧化碳还原氧化铁（III）提取：$\mathrm{Fe_2O_3(s) + 3\,CO(g) \to 2\,Fe(s) + 3\,CO_2(g)}$。一座冶炼厂投入 $160.0\ \mathrm{g}$ $\mathrm{Fe_2O_3}$ 和 $84.0\ \mathrm{g}$ CO，实际铁产量为 $97.0\ \mathrm{g}$。（$M_{\mathrm{Fe_2O_3}} = 159.69$，$M_{\mathrm{CO}} = 28.01$，$M_{\mathrm{Fe}} = 55.85\ \mathrm{g\,mol^{-1}}$）
(a)Calculate the moles of each reactant.计算每种反应物的摩尔数。[2]
(b)Determine the limiting reactant with a clear comparison. State which reactant is in excess.通过清晰的比较确定限量反应物，并说明哪种反应物过量。[3]
(c)Calculate the theoretical yield of iron in grams.计算铁的理论产量（以克为单位）。[2]
(d)Calculate the percent yield of iron.计算铁的产率。[2]
Q12HARD难🇺🇸 US美AP-feeder FRQAP 衔接简答题§2 + §3 + §5 Full stoichiometry chain: combustion analysis to product mass完整化学计量链：从燃烧分析到产物质量[8 marks][8 分]
(a)Calculate the number of moles of ethanol in the sample.计算样品中乙醇的摩尔数。[1]
(b)Calculate the mass of $\mathrm{CO_2}$ produced. ($M_{\mathrm{CO_2}} = 44.01\ \mathrm{g\,mol^{-1}}$)计算生成 $\mathrm{CO_2}$ 的质量。（$M_{\mathrm{CO_2}} = 44.01\ \mathrm{g\,mol^{-1}}$）[3]
(c)Calculate the mass of $\mathrm{H_2O}$ produced. ($M_{\mathrm{H_2O}} = 18.02\ \mathrm{g\,mol^{-1}}$)计算生成 $\mathrm{H_2O}$ 的质量。（$M_{\mathrm{H_2O}} = 18.02\ \mathrm{g\,mol^{-1}}$）[2]
(d)Verify your answers to (b) and (c) using conservation of mass. Show the check clearly.用质量守恒定律验证 (b) 和 (c) 的答案，清晰写出验证过程。[2]
🇺🇸 US NGSS美国 NGSSHS-PS1-7mole ratios, stoichiometry, limiting reactants摩尔比、化学计量、限量反应物