Course: 12.3 质量-最低存储测量法

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质量- 最低口式测量法
Need nails?
::需要钉子吗?

When you are doing a large construction project, you have a good idea of how many nails you will need (lots!). When you go to the hardware store, you don’t want to sit there and count out several hundred nails. You can buy nails by weight , so you determine how many nails are in a pound, calculate how many pounds you need, and you’re on your way to begin building.
::当您正在做一个大型建筑工程时,您可以很好地知道需要多少钉子( lots! ) 。当你去五金店时, 您不想坐在那里点数几百个钉子。您可以按重量购买钉子, 这样您就可以确定每磅有多少钉子, 算出您需要多少钉子, 并且您正准备开始建造。

While the is ever-present in all stoichiometry calculations, amounts of substances in the laboratory are most often measured by mass. Therefore, we need to use mole-mass calculations in combination with mole ratios to solve several different types of mass-based stoichiometry problems.
::虽然在所有的测量方法计算中始终存在这种物质,但实验室中物质的数量通常以质量衡量。因此,我们需要结合摩尔比计算摩尔质量,以解决几种不同类型的基于质量的测量方法问题。

Mass to Moles Problems
::质量到摩尔问题

In this type of problem, the mass of one substance is given, usually in grams. From this, you are to determine the amount in moles of another substance that will either react with or be produced from the given substance.
::在这类问题中,给定一种物质的质量,通常以克为单位。从这一点上,你将确定另一种物质与给定物质发生反应或由特定物质产生的数量。

$\begin{aligned} mass of g i v e n \to moles of g i v e n \to moles of u n k n o w n \end{aligned}$
$\begin{align*}\text{mass of} \ given \rightarrow \text{moles of} \ given \rightarrow \text{moles of} \ unknown\end{align*}$
::未知的给定摩擦

The mass of the given substance is converted into moles by use of the of that substance from the periodic table . Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation .
::特定物质的质量通过使用周期表中的该物质而转换成摩尔。然后,特定物质的摩尔通过使用平衡化学方程的摩尔比转换成未知的摩尔。

Sample Problem: Mass-Mole Stoichiometry
::样本问题: 质量- 摩尔存储测量法

Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen according to the following balanced equation .
::锡金属与氟化氢发生反应,根据以下平衡方程生产锡(II)氟化物和氢。

$\begin{aligned} Sn (s) + 2 HF (g) \to {SnF}_{2} (s) + H_{2} (g) \end{aligned}$
$\begin{align*}\text{Sn}(s)+2\text{HF}(g) \rightarrow \text{SnF}_2(s)+\text{H}_2 (g)\end{align*}$
::Sn(s)+2HF(g)+SnF2(s)+H2(g)

How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?
::用75.0克锡完全反应需要多少个氟化氢的摩尔?

Step 1: List the known quantities and plan the problem.
::第1步:列出已知数量并规划问题。

Known
::已知已知

given: 75.0 g Sn
::给定: 75.0 g Sn

molar mass of Sn = 118.69 g/mol
::Sn = 118.69克/摩尔的摩尔质量

1 mol Sn = 2 mol HF (mole ratio)
::1 mol Sn = 2 mol HF(死亡率)

Unknown
::未知

mol HF
::mol Mol 高频

Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.
::使用 Sn 的摩尔质量将 Sn 克转换为摩尔。然后使用摩尔比将摩尔 Sn 转换为摩尔高频。这将在单两步计算中完成。

g Sn → mol Sn → mol HF
::g ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Step 2: Solve.
::步骤2:解决。

$\begin{aligned} 75.0 g Sn \times \frac{1 mol Sn}{118.69 g Sn} \times \frac{2 mol HF}{1 mol Sn} = 1.26 mol HF \end{aligned}$
$\begin{align*}75.0 \text{ g Sn} \times \frac{1 \text{ mol Sn}}{118.69 \text{ g Sn}} \times \frac{2 \text{ mol HF}}{1 \text{ mol Sn}}=1.26 \text{ mol HF}\end{align*}$
::75.0克 Snx1 mol Sn118.69克 Snx2 mol HF1 mol Sn=1.26 mol HF

Step 3: Think about your result.
::步骤3:想想你的结果。

The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three because the given mass has three significant figures.
::锡的质量小于一个摩尔,但1:2的比例表示反应需要超过一个摩尔的HF。答案有三个,因为特定质量有三个重要数字。

Moles to Mass Problems
::与大众问题混合

In this type of problem, the amount of one substance is given in moles. From this, you are to determine the mass of another substance that will either react with or be produced from the given substance.
::在这类问题中,一种物质的数量以摩尔形式给出。从这一点上,你将确定另一种物质的质量,这种物质要么与给定物质发生反应,要么由给定物质产生。

$\begin{aligned} moles of g i v e n \to moles of u n k n o w n \to mass of u n k n o w n \end{aligned}$
$\begin{align*}\text{moles of} \ given \rightarrow \text{moles of} \ unknown \rightarrow \text{mass of} \ unknown\end{align*}$
::未知的未知质量

The moles of the given substance are first converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Then, the moles of the unknown are converted into mass in grams by use of the molar mass of that substance from the periodic table.
::使用平衡化学方程式的摩尔比,先将特定物质的摩尔转化为未知物质的摩尔。然后,利用周期表中该物质的摩尔质量,将未知物质的摩尔比转换成质量(克)。

Sample Problem: Mole-Mass Stoichiometry
::样本问题: 分子- 混集层测量法

Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor .
::氧气中硫化氢气体燃烧,产生二氧化硫和水蒸气。

$\begin{aligned} 2 H_{2} S (g) + 3 O_{2} (g) \to 2 {SO}_{2} (g) + 2 H_{2} O (g) \end{aligned}$
$\begin{align*}2\text{H}_2 \text{S}(g) + 3\text{O}_2(g) \rightarrow 2\text{SO}_2(g)+2\text{H}_2\text{O}(g)\end{align*}$
::2H2S(g)+3O2(g)%2SO2(g)+2H2O(g)

What mass of oxygen gas is consumed in a reaction that produces 4.60 mol SO ₂ ?
::产生4.60 摩尔二氧化硫的反应中消耗的氧气质量是多少?

Step 1: List the known quantities and plan the problem.
::第1步:列出已知数量并规划问题。

Known
::已知已知

given: 4.60 mol SO ₂
::说明:4.60 毫微升

2 mol SO ₂ = 3 mol O ₂ (mole ratio)
::2 mol SO2 = 3 mol O2(死亡率)

molar mass of O ₂ = 32.00 g/mol
::O2 = 32.00克/摩尔摩尔

Unknown
::未知

mass O ₂ = ? g
::质量 O2 = ? g

Use the mole ratio to convert from mol SO ₂ to mol O ₂ . Then convert mol O ₂ to grams. This will be done in a single two-step calculation.
::使用摩尔比将摩尔SO2转换为摩尔O2,然后将摩尔O2转换为克。这将在单步两步计算中完成。

mol SO ₂ → mol O ₂ → g O ₂
::mol SO2 mol O2 g O2

Step 2: Solve.
::步骤2:解决。

$\begin{aligned} 4.60 {mol SO}_{2} \times \frac{3 {mol O}_{2}}{2 {mol SO}_{2}} \times \frac{32.00 {g O}_{2}}{1 {mol O}_{2}} = 221 {g O}_{2} \end{aligned}$
$\begin{align*}4.60 \text{ mol SO}_2 \times \frac{3 \text{ mol O}_2}{2 \text{ mol SO}_2} \times \frac{32.00 \text{ g O}_2}{1 \text{ mol O}_2}=221 \text{ g O}_2\end{align*}$
::4.60 mol SO2×3 mol O22 mol SO2×32.00 g O21 mol O2=221 g O2

Step 3: Think about your result.
::步骤3:想想你的结果。

According to the mole ratio, 6.90 mol O ₂ is consumed with a mass of 221 g. The answer has three significant figures because the given number of moles has three significant figures.
::根据摩尔比率,6.90毫升O2的消耗质量为221克。答案有三个重要数字,因为给定的摩尔数量有三个重要数字。

Summary
::摘要

Calculations involving conversions of mass to moles and moles to mass are described.
::说明涉及将质量转化为摩尔和摩尔转化为质量的计算。

Review
::回顾

In the first problem, what would happen if you multiply grams Sn by 118.69 grams/mole Sn?
::第一个问题,如果你把Sn克乘以118.69克/摩尔Sn,会发生什么?

Why is a balanced equation needed?
::为什么需要平衡的等式?

Does the physical form of the material matter for these calculations?
::这些计算是否与物质的物质形式有关?

Section outline

General

质量- 最低口式测量法

Need nails? ::需要钉子吗?

Mass to Moles Problems ::质量到摩尔问题

Sample Problem: Mass-Mole Stoichiometry ::样本问题: 质量- 摩尔存储测量法

Moles to Mass Problems ::与大众问题混合

Sample Problem: Mole-Mass Stoichiometry ::样本问题: 分子- 混集层测量法

Summary ::摘要

Review ::回顾

Need nails?
::需要钉子吗?

Mass to Moles Problems
::质量到摩尔问题

Sample Problem: Mass-Mole Stoichiometry
::样本问题: 质量- 摩尔存储测量法

Moles to Mass Problems
::与大众问题混合

Sample Problem: Mole-Mass Stoichiometry
::样本问题: 分子- 混集层测量法

Summary
::摘要

Review
::回顾