Reacting masses and yield · 反应物质量与产率
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| mole ratio/məʊl ˈreɪʃɪəʊ/ | 摩尔比 | mó ěr bǐ |
| stoichiometry/ˌstəʊɪkɪˈɒmətri/ | 化学计量 | huà xué jì liàng |
| percentage yield/pəˈsentɪdʒ jiːld/ | 产率 | chǎn lǜ |
| excess reagent/ekˈses rɪˈeɪdʒənt/ | 过量试剂 | guò liàng shì jì |
| limiting reagent/ˈlɪmɪtɪŋ rɪˈeɪdʒənt/ | 限量试剂 | xiàn liàng shì jì |
How much product will you get?
- The numbers in a balanced equation give the mole ratio 摩尔比 — the stoichiometry 化学计量.
- From it we work out reacting masses.
- Real reactions give less than the maximum, and one reactant runs out first.
你能得到多少产物?
- 配平方程式中的数字给出 摩尔比(mole ratio)——化学计量(stoichiometry)。
- 由它我们算出反应质量。
- 真实的反应给出比最大值少的量,而且一种反应物先用完。
Reacting mass route · 反应物质量路径
Follow a balanced equation from known mass to predicted product mass. · 根据配平的化学方程式，从已知质量推算产物质量。
Percentage yield is the actual yield divided by the theoretical yield, times 100. · 产率等于实际产量除以理论产量，再乘以 100。
It measures how much product you actually obtained compared with the maximum possible. · 它衡量你实际获得的产物量与最大可能量的对比。
The stoichiometry of a reaction comes from: · 反应的化学计量数来自：
The balancing numbers give the mole ratio in which species react and form. · 配平系数给出了反应物之间形成产物的摩尔比。
The reacting-mass method
To find a reacting mass:
- change the known mass → moles ($n = m/M$).
- use the mole ratio to find the moles you want.
- change those moles → mass.
反应质量法
求一个反应质量:
- 把已知 质量 → 摩尔数($n = m/M$)。
- 用 摩尔比 求出你想要的摩尔数。
- 把那些 摩尔数 → 质量。
Put the steps of a reacting-mass calculation in order. · 将反应质量计算的步骤按正确顺序排列。
mass → moles → ratio → moles → mass. The mole ratio is the bridge between the two substances. · 质量 → 摩尔数 → 比例 → 摩尔数 → 质量。摩尔比是连接两种物质的桥梁。
In 2H₂ + O₂ → 2H₂O, how many moles of O₂ are needed to react with 4 mol of H₂? · 在 2H₂ + O₂ → 2H₂O 中，与 4 mol H₂ 反应需要多少摩尔的 O₂？
The ratio H₂ : O₂ is 2 : 1, so 4 mol H₂ needs 4 ÷ 2 = 2 mol O₂. · H₂ : O₂ 的比例为 2 : 1，因此 4 mol H₂ 需要 4 ÷ 2 = 2 mol O₂。
Percentage yield 产率
- You usually get less product than the maximum.
百分产率
- 你通常得到比最大值 少 的产物。
A reaction makes 8 g of product when the maximum possible is 10 g. What is the percentage yield? · 某反应生成了 8 g 产物，而最大可能产量为 10 g。求产率百分比？
percentage yield = (actual / maximum) × 100 = (8 / 10) × 100 = 80%. · 产率 = (实际 / 最大) × 100 = (8 / 10) × 100 = 80%。
Limiting and excess reagent 过量试剂
- The limiting reagent 限量试剂 runs out first and decides how much product forms.
- The excess reagent is left over.
- To find it: divide each reactant's moles by its number in the equation — the smallest result is limiting.
The reactant that runs out first (the limiting reagent) sets how much product forms.
限量试剂和过量试剂
- 限量试剂(limiting reagent) 先用完,决定 生成多少产物。
- 过量试剂(excess reagent) 剩下来。
- 找它的方法:把每种反应物的摩尔数除以它在方程式中的数目——最小的 结果是限量的。
先用完的反应物(限量试剂)决定生成多少产物。
The limiting reagent in a reaction is the one that: · 反应中的限量试剂是：
The limiting reagent is used up first, so it sets the amount of product; the other is in excess. · 限量试剂最先被消耗完，因此它决定了产物的量；另一种物质则是过量的。
The reactant that runs out first and decides how much product forms is the ______ reagent. · 最先耗尽并决定产物生成量的反应物是 ______ 试剂。
The other reactant is "in excess" — some of it is left over. · 另一种反应物处于“过量”状态——其中一部分会剩余。
You've got it
- stoichiometry = mole ratio from the balanced equation
- reacting mass: mass → moles → ratio → moles → mass
- $\text{percentage yield} = \dfrac{\text{actual}}{\text{maximum}} \times 100\%$
- the limiting reagent runs out first and sets the product; base the calculation on it
你掌握了
- 化学计量 = 来自配平方程式的摩尔比
- 反应质量:质量 → 摩尔数 → 比 → 摩尔数 → 质量
- $\text{percentage yield} = \dfrac{\text{actual}}{\text{maximum}} \times 100\%$
- 限量试剂 先用完并决定产物;计算以它为基准