Stoichiometry · 化学计量学
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| mole ratio/məʊl ˈreɪʃɪəʊ/ | 摩尔比 | mó ěr bǐ |
| limiting reactant/ˈlɪmɪtɪŋ rɪˈæktənt/ | 限量反应物 | xiàn liàng fǎn yìng wù |
The recipe math of chemistry
- A recipe says 2 eggs per cake -- chemistry has ratios too.
- A balanced equation gives the exact mole proportions.
- From how much you start with, you predict how much you will make.
- It is the arithmetic that runs every reaction.
化学的食谱算术
- 食谱说每个蛋糕 2 个鸡蛋——化学也有比例。
- 配平的方程给出精确的摩尔比例。
- 从你起始的量,你能预测将生成多少。
- 这是驱动每个反应的算术。
Mole ratios from the equation
- The coefficients give the mole ratio 摩尔比 between substances.
- In $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$, that is 2 mol $\text{H}_2$ per 1 mol $\text{O}_2$.
- Use that ratio to convert moles of one into moles of another.
从方程得出摩尔比
- 系数给出物质之间的摩尔比。
- 在 $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$ 中,即每 1 摩尔 $\text{O}_2$ 对 2 摩尔 $\text{H}_2$。
- 用这个比把一种物质的摩尔数换算成另一种的。
In $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$, how many moles of water form from 3 mol of $\text{O}_2$ (excess $\text{H}_2$)? · 在 $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$ 中，从 3 mol 的 $\text{O}_2$（过量 $\text{H}_2$）生成多少摩尔的水？
The ratio is 1 $\text{O}_2$ to 2 $\text{H}_2\text{O}$, so $3 \times 2 = 6\ \text{mol}$. · 比例为 1 $\text{O}_2$ 比 2 $\text{H}_2\text{O}$，所以是 $3 \times 2 = 6\ \text{mol}$。
You must balance the equation before reading off the mole ratio. · 读取摩尔比之前必须先配平方程式。
The coefficients only give the correct ratio when balanced. · 只有在配平后，系数才给出正确的比例。
Grams to grams
- Convert grams to moles (divide by molar mass), apply the ratio, convert back.
- Grams, then moles, then moles, then grams -- the master path.
- Every stoichiometry problem walks this road.
从克到克
- 把克换算成摩尔(除以摩尔质量),用比例,再换算回去。
- 克,再摩尔,再摩尔,再克——这条主路径。
- 每道化学计量题都走这条路。
Order the steps of a grams-to-grams stoichiometry problem. · 对克到克的化学计量问题步骤进行排序。
Grams to moles, apply ratio, moles back to grams. · 克转摩尔，应用比例，摩尔再转回克。
How many grams is 2 mol of water (molar mass 18)? · 2 mol 水的质量是多少（摩尔质量为 18）？
$m = nM = 2 \times 18 = 36\ \text{g}$.
The limiting reactant
- The limiting reactant 限量反应物 runs out first and caps the product.
- Compare the available moles against the ratio to find it.
- Whatever is left over is the excess reactant.
限量反应物
- 限量反应物先耗尽,并限定产物的量。
- 把可用的摩尔数与比例对照,找出它。
- 剩下的就是过量反应物。
The stoichiometry roadmap · 化学计量学路线图
Convert a mass of one substance to a mass of another using the balanced equation. · 利用配平的方程式将一种物质的质量转换为另一种物质的质量。
The limiting reactant is the one that... · 限量反应物是那个...
The reactant that runs out first caps how much product can form. · 最先耗尽的反应物决定了能形成多少产物。
The reactant left over after the reaction stops is called the ____ reactant. · 反应停止后剩余的反应物被称为 ____ 反应物。
The non-limiting reactant is in excess. · 非限量反应物处于过量状态。
How many moles of $\text{H}_2\text{O}$ form from $4\ \text{mol}$ of $\text{H}_2$ (with plenty of $\text{O}_2$)?
- The ratio $2\text{H}_2 \to 2\text{H}_2\text{O}$ is 1 : 1.
- So $4\ \text{mol}$ of $\text{H}_2$ gives $4\ \text{mol}$ of $\text{H}_2\text{O}$.
$4\ \text{mol}$ 的 $\text{H}_2$(有充足的 $\text{O}_2$)生成多少摩尔 $\text{H}_2\text{O}$?
- 比例 $2\text{H}_2 \to 2\text{H}_2\text{O}$ 是 1 : 1。
- 所以 $4\ \text{mol}$ 的 $\text{H}_2$ 生成 $4\ \text{mol}$ 的 $\text{H}_2\text{O}$。
Always start from a balanced equation -- the coefficients are the mole ratio. Convert masses to moles before using the ratio, because you cannot compare grams directly. And the limiting reactant, not the bigger pile, sets how much product forms.
始终从配平的方程开始——系数就是摩尔比。在使用比例之前把质量换算成摩尔,因为你不能直接比较克数。而且是限量反应物、而不是更大的那一堆,决定生成多少产物。
Stoichiometry uses the balanced equation's mole ratio to predict amounts. Convert grams to moles, apply the ratio, then convert back to grams. When reactants are both limited, the limiting reactant runs out first and caps the product; the rest is in excess.
化学计量用配平方程的摩尔比来预测量。把克换算成摩尔,用比例,再换算回克。当两种反应物都有限时,限量反应物先耗尽并限定产物;其余为过量。