节: 批量口测量法 | 12.6 大规模蒸汽测量

章节大纲

How much azide is needed to fill an air bag?
::填充空气袋需要多少安眠药?

Cars and many other vehicles have air bags in them. In case of a collision, a reaction is triggered so that the rapid decomposition of sodium azide produces nitrogen , filling the air bag. If too little sodium azide is used, the air bag will not fill completely and will not protect the person in the vehicle. Too much sodium azide could cause the formation of more gas that the bag can safely handle. If the bag breaks from the excess , all protection is lost.
::汽车和许多其他车辆都装有气袋。发生碰撞时,会触发反应, 使Azide钠迅速分解产生氮, 填满气袋。如果使用过少的Azide钠, 气袋不会完全填满, 也不会保护车上的人。太多的Azide钠可能会形成更多的气体, 气袋可以安全地处理。如果气袋从多余的气袋中解开, 所有的保护就会丢失。

Mass to Volume and Volume to Mass Problems
::质量到量到量到量量到量到量到量量到量到量到量问题

frequently involve both solid substances whose mass can be measured as well as gases for which measuring the volume is more appropriate. Stoichiometry problems of this type are called either mass-volume or volume-mass problems.
::通常既涉及其质量可以测量的固体物质,也涉及测量其体积更适当的气体。

$\begin{aligned} mass of g i v e n \to moles of g i v e n \to moles of u n k n o w n \to volume of u n k n o w n \\ volume of g i v e n \to moles of g i v e n \to moles of u n k n o w n \to mass of u n k n o w n \end{aligned}$
$\begin{align*}& \text{mass of} \ given \ \rightarrow \ \text{moles of} \ given \ \rightarrow \ \text{moles of} \ unknown \ \rightarrow \ \text{volume of} \ unknown\\ & \text{volume of} \ given \ \rightarrow \ \text{moles of} \ given \ \rightarrow \ \text{moles of} \ unknown \ \rightarrow \ \text{mass of} \ unknown\end{align*}$
::给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 给定 moles 未知 moles 未知 moles 数量为未知数量未知 moles 未知 moles 未知 moles

Because both types of problems involve a conversion from either moles of gas to volume or vice-versa, we can use the molar volume of 22.4 L/mol provided that the conditions for the reaction are STP.
::由于这两类问题都涉及将气体的摩尔转化为体积或反转体,我们可以使用22.4升/摩尔的摩尔体积,只要反应的条件是保护受威胁人民协会。

Sample Problem: Mass-Volume Stoichiometry
::样本问题: 大规模蒸汽蒸汽测量

Aluminum metal reacts rapidly with aqueous sulfuric to produce aqueous aluminum sulfate and hydrogen gas.
::铝金属与水硫反应迅速,产生水铝硫酸盐和氢气。

$\begin{aligned} 2 Al (s) + 3 H_{2} {SO}_{4} (a q) \to {Al}_{2} ({SO}_{4})_{3} (a q) + 3 H_{2} (g) \end{aligned}$
$\begin{align*}2 \text{Al}(s)+3\text{H}_2\text{SO}_4 (aq) \rightarrow \text{Al}_2 (\text{SO}_4)_3 (aq)+3\text{H}_2 (g)\end{align*}$
::2Al(s)+3H2SO4(aq) = AL2(Al2(SO4)3(aq)+3H2(g)

Determine the volume of hydrogen gas produced at STP when a 2.00 g piece of aluminum completely reacts.
::2.00克铝块完全反应时,确定在STTP产生的氢气的量。

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: 2.00 g Al
::说明:2.00g Al

Al = 26.98 g/mol
::阿尔=26.98克/摩尔

2 mol Al = 3 mol H ₂
::2毫升=3毫升H2

Unknown
::未知

volume H ₂ = ?
::体积H2=?

The grams of aluminum will first be converted to moles. Then the will be applied to convert to moles of hydrogen gas. Finally, the molar volume of a gas will be used to convert to liters of hydrogen.
::铝克将首先转换成摩尔。然后, 将应用于转换成氢气的摩尔。最后, 气体的摩尔体积将被用于转换成氢升。

$\begin{aligned} g Al \to mol Al \to {mol H}_{2} \to {L H}_{2} \end{aligned}$
$\begin{align*}\text{g Al} \rightarrow \text{mol Al} \rightarrow \text{mol H}_2 \rightarrow \text{L H}_2\end{align*}$
::g Al-mol Al-mol H2L H2

Step 2: Solve.
::步骤2:解决。

$\begin{aligned} 2.00 g Al \times \frac{1 mol Al}{26.98 g Al} \times \frac{3 {mol H}_{2}}{2 mol Al} \times \frac{22.4 {L H}_{2}}{1 {mol H}_{2}} = 2.49 {L H}_{2} \end{aligned}$
$\begin{align*}2.00 \text{ g Al} \times \frac{1 \text{ mol Al}}{26.98 \text{ g Al}} \times \frac{3 \text{ mol H}_2}{2 \text{ mol Al}} \times \frac{22.4 \text{ L H}_2}{1 \text{ mol H}_2} = 2.49 \text{ L H}_2\end{align*}$
::2.00克 2.00克 2.00克 2.8克 2.6.98克 2.3 摩摩 H22 摩摩 ALx 22.4 L H21 摩摩 H2= 2.49 L H2

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

The volume result is in liters. For much smaller amounts, it may be convenient to convert to milliliters. The answer here has three . Because the molar volume is a measured quantity of 22.4 L/mol, three is the maximum number of significant figures for this type of problem.
::体积结果为升数。对于小得多的数量, 转换为毫升可能比较方便。答案在这里有三个。因为摩尔体积的测量量为22.4升/ 摩尔, 3 是这类问题的重要数字的最大数量。

Sample Problem: Volume-Mass Stoichiometry
::样本问题: 量- 量- 量 - 量 - 量 - 量 - 量 - 量 - 量 - 度

Calcium oxide is used to remove sulfur dioxide generated in coal-burning power plants according to the following reaction.
::氧化钙用于根据以下反应清除燃煤发电厂产生的二氧化硫。

$\begin{aligned} 2 CaO (s) + 2 {SO}_{2} (g) + O_{2} (g) \to 2 {CaSO}_{4} (s) \end{aligned}$
$\begin{align*}2\text{CaO}(s) + 2\text{SO}_2 (g)+\text{O}_2 (g) \rightarrow 2\text{CaSO}_4 (s)\end{align*}$
::2CaO+2SO2(g)+O2(g)+O2(g)%2CASO4(s)

What mass of calcium oxide is required to react completely with 1.4 × 10 ³ L of sulfur dioxide?
::用1.4×103升的二氧化硫完全反应需要多大质量的氧化钙?

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: 1.4 × 10 ³ L = SO ₂
::说明: 1.4 × 103 L = SO2

2 mol SO ₂ = 2 mol CaO
::2 mol SO2 = 2 mol CAO

molar mass CaO = 56.08 g/mol
::摩尔质量CAO=56.08克/摩尔

Unknown
::未知

mass CaO = ? g
::质量 CAO = ? g

The volume of SO ₂ will be converted to moles, followed by the mole ratio, and finally a conversion of moles of CaO to grams.
::SO2的体积将转换成摩尔,然后是摩尔比率,最后是CAO的摩尔转换成克。

$\begin{aligned} {L SO}_{2} \to {mol SO}_{2} \to mol CaO \to g CaO \end{aligned}$
$\begin{align*}\text{L SO}_2 \rightarrow \text{mol SO}_2 \rightarrow \text{mol CaO} \rightarrow \text{g CaO}\end{align*}$
::CAOEG CAO 氧化碳(CAO)

Step 2: Solve.
::步骤2:解决。

$\begin{aligned} 1.4 \times 10^{3} {L SO}_{2} \times \frac{1 {mol SO}_{2}}{22.4 {L SO}_{2}} \times \frac{2 mol CaO}{2 {mol SO}_{2}} \times \frac{56.08 g CaO}{1 mol CaO} = 3.5 \times 10^{3} g CaO \end{aligned}$
$\begin{align*}1.4 \times 10^3 \text{ L SO}_2 \times \frac{1 \text{ mol SO}_2}{22.4 \text{ L SO}_2} \times \frac{2 \text{ mol CaO}}{2 \text{ mol SO}_2} \times \frac{56.08 \text{ g CaO}}{1 \text{ mol CaO}}=3.5 \times 10^3 \text{ g CaO}\end{align*}$
::1.4×103 L SO2×1 mol SO222.4 L SO2×2 mol CaO2 mol CaO2 mol SO2×56.08 g CAO1 mol CaO=3.5×103 g CAO

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

The resultant mass could be reported as 3.5 kg, with two significant figures. Even though the 2:2 mole ratio does not mathematically affect the problem, it is still necessary for unit conversion.
::由此产生的质量可报告为3.5公斤,其中有两个重要数字,尽管2:2摩尔比率在数学上并不影响这一问题,但单位转换仍有必要。

Summary
::摘要

Calculations are described for determining the amount of gas formed in a reaction.
::用于确定反应中形成的气体数量的计算方法说明。

Calculations are described for determining amounts of a material needed to react with a gas.
::为确定对气体作出反应所需材料的数量而作了计算。

Review
::回顾

What are the conditions for all gases in these calculations?
::这些计算中所有气体的条件是什么?

How can you tell if all the ratios were set up correctly?
::你如何判断所有比率是否都正确?

Why was 2 mol CaO/2mol SO ₂ included in the second example if it did not affect the final number?
::为什么第二个例子中包括2 mol CaO/2mol SO2,如果它不影响最后数字的话?

章节大纲

How much azide is needed to fill an air bag? ::填充空气袋需要多少安眠药?

Mass to Volume and Volume to Mass Problems ::质量到量到量到量 量到量到量到量 量到量到量到量问题

Sample Problem: Mass-Volume Stoichiometry ::样本问题: 大规模蒸汽蒸汽测量

Sample Problem: Volume-Mass Stoichiometry ::样本问题: 量- 量- 量 - 量 - 量 - 量 - 量 - 量 - 量 - 度

Summary ::摘要

Review ::回顾

How much azide is needed to fill an air bag?
::填充空气袋需要多少安眠药?

Mass to Volume and Volume to Mass Problems
::质量到量到量到量量到量到量到量量到量到量到量问题

Sample Problem: Mass-Volume Stoichiometry
::样本问题: 大规模蒸汽蒸汽测量

Sample Problem: Volume-Mass Stoichiometry
::样本问题: 量- 量- 量 - 量 - 量 - 量 - 量 - 量 - 量 - 度

Summary
::摘要

Review
::回顾