📚 Mastering Calculation Questions in AQA GCSE Chemistry | 掌握AQA GCSE化学计算题型
In AQA GCSE Chemistry, calculation questions are a significant part of the exams and require a solid understanding of quantitative chemistry concepts. Mastering these calculations can boost your confidence and grade. This article covers the essential calculation types found in the AQA GCSE Chemistry Student Book, providing clear explanations and step-by-step approaches.
在AQA GCSE化学考试中，计算题占据了重要部分，需要扎实掌握定量化学概念。熟练掌握这些计算能够提升你的信心和成绩。本文涵盖AQA GCSE化学学生用书中出现的核心计算题型，提供清晰的解释与分步方法。
1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对分子质量
r found on the periodic table. Relative formula mass (Mᵣ) is the sum of the Aᵣ values of all atoms in a chemical formula. For example, Mᵣ of H₂O = 2 × 1 + 16 = 18. For ionic compounds such as NaCl, we also use the term relative formula mass, which is calculated similarly.
相对原子质量(Aᵣ)是元素原子平均质量与碳-12原子质量的1/12的比值，为无量纲数，可从周期表中获取。相对分子质量(Mᵣ)是化学式中所有原子Aᵣ之和。例如，H₂O的Mᵣ = 2×1 + 16 = 18。对于离子化合物如NaCl，我们也使用相对式量这一术语，计算方式相同。
40, C = 12, O₃ = 3×16 = 48, so Mᵣ = 40 + 12 + 48 = 100. Always show your working clearly.
常见的考试任务是计算给定化学式的Mᵣ。对于CaCO₃：Ca = 40，C = 12，O₃ = 3×16 = 48，因此Mᵣ = 40 + 12 + 48 = 100。务必清晰地展示计算过程。
2. The Mole and Molar Mass | 摩尔与摩尔质量
vogadro constant). The molar mass (in g/mol) is numerically equal to the relative formula mass (Mᵣ). For example, the molar mass of H₂O is 18 g/mol. The key formula linking mass and moles is:
摩尔是物质的量的单位。1摩尔任何物质恰好含有6.02×10²³个粒子（阿伏伽德罗常数）。摩尔质量（单位g/mol）在数值上等于相对分子质量(Mᵣ)。例如，H₂O的摩尔质量是18 g/mol。连接质量和摩尔数的关键公式为：
Number of moles (n) = mass (m) ÷ Mᵣ
rranged to find mass = n × Mᵣ or Mᵣ = mass / n. Practise using it in different situations.
该公式可变形为质量 = n × Mᵣ 或 Mᵣ = 质量 / n。请在各种情境下练习使用此公式。
3. Reacting Mass Calculations | 反应质量计算
To calculate the mass of a product formed or reactant needed, follow these steps: write the balanced equation, convert given mass to moles, use the mole ratio from the equation to find moles of the target substance, then convert back to mass.
计算生成物的质量或所需反应物的质量时，按以下步骤操作：写出配平的化学方程式，将已知质量转换为摩尔数，利用方程式中的摩尔比求出目标物质的摩尔数，再转换回质量。
Example: What mass of MgO is formed when 6 g of Mg burns completely? ole ratio 1:1. Moles of MgO = 0.25 mol. Mᵣ of MgO = 24 + 16 = 40. Mass = 0.25 × 40 = 10 g.
示例：6 g镁完全燃烧生成多少克MgO？2Mg + O₂ → 2MgO。Mg的摩尔数 = 6/24 = 0.25 mol。由方程式，2 mol Mg生成2 mol MgO，摩尔比为1:1。MgO的摩尔数 = 0.25 mol。MgO的Mᵣ = 24+16 = 40。质量 = 0.25 × 40 = 10 g。
4. Limiting Reactants | 限制反应物
In a reaction, the reactant that is completely used up first is the limiting reactant; it determines the amount of product formed. The other reactant is in excess. To identify the limiting reactant, calculate the moles of each reactant, then compare the mole ratio from the balanced equation. The reactant that gives the smaller number of moles (after adjusting for the stoichiometric ratio) is limiting.
在反应中，首先被完全消耗的反应物是限制反应物，它
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