14.9 气体摩尔质量计算

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  计算气体的摩尔质量

  

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  What makes it float?
  ::是什么让它浮起来的?

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  Helium has long been used in balloons and blimps. Since it is much less dense than air, it will float above the ground. We can buy small balloons filled with helium at stores, but large ones (such as the balloon seen above) are much more expensive and take up a lot more helium.
  ::长期以来一直被用于气球和浮质中。 由于它比空气密度低得多,它会漂浮在地面上。 我们可以在商店购买装满的小型气球,但大型气球(如上面所看到的气球)的成本要高得多,而且吸收了更多的。

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  Calculating Molar Mass and Density of a Gas
  ::计算气体的摩尔质量和密度

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  A , which produces a , is performed. The produced gas is then collected and its mass and volume are determined. The of the unknown gas can be found using the , provided the temperature and pressure of the gas are also known.
  ::产生 a 的 A 进行,然后收集产生的气体并确定其质量和体积。使用该气体可以找到未知气体的重量和体积,前提是该气体的温度和压力也为人所知。

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  Sample Problem: Molar Mass and the Ideal Gas Law
  ::抽样问题:摩尔弥撒和理想气体法

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  A certain reaction occurs, producing an oxide of nitrogen as a gas. The gas has a mass of 1.211 g and occupies a volume of 677 mL. The temperature in the laboratory is 23°C and the air pressure is 0.987 atm. Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal.
  ::气体质量为1.211克,体积为677毫升。实验室温度为23°C,空气压力为0.987 原子。计算气体的摩尔质量并推断其公式。假设气体是理想的。

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

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  Known
  ::已知已知

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  • $\begin{align*}\text{mass}=1.211\text{ g}\end{align*}$
    ::质量=1.211克
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  • $\begin{align*}V = 677 \text{ ml} = 0.677 \text{ L}\end{align*}$
    ::V=677毫升=0.677升
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  • $\begin{align*}T= 23^\circ \text{C}=296 \text{ K}\end{align*}$
    ::T=23°C=296K
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  • $\begin{align*}P =0.987 \ \text{atm}\end{align*}$
    ::P=0.987 小时
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  • $\begin{align*}R =0.08206 \text{ L} \cdot \text{atm/K} \cdot \text{mol}\end{align*}$
    ::R=0.08206里拉特姆/克摩尔

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  Unknown
  ::未知

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  • $\begin{align*}n= ? \ \text{mol}\end{align*}$
    ::不=? 摩尔
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  • $\begin{align*}\text{molar mass} = ? \ \text{g/mol}\end{align*}$
    ::摩尔质量=? g/摩尔

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  First the ideal gas law will be used to solve for the moles of unknown gas "> $\begin{align*}(n)\end{align*}$ . Then the mass of the gas divided by the moles will give the molar mass.
  ::首先,理想的天然气法将被用于解决未知气体的摩尔 。 然后,由摩尔分子分开的气体质量将产生摩尔质量。

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  Step 2: Solve .
  ::步骤2:解决。

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  $$\begin{align*}n=\frac{PV}{RT}=\frac{0.987 \ \text{atm} \times 0.677 \text{ L}}{0.08206 \text{ L} \cdot \text{atm/K} \cdot \text{mol} \times 296 \text{ K}}=0.0275 \ \text{mol}\end{align*}$$
  ::n= PVRT=0.987 atmx0.677 L.08206 Latm/Kmolx296 K=0.0275 毫升

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  Now divide g by mol to get the molar mass.
  ::现在按摩尔和摩尔分开 以获得摩尔质量

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  $$\begin{align*}\text{molar mass}=\frac{1.211 \text{ g}}{0.0275 \ \text{mol}}=44.0 \ \text{g/mol}\end{align*}$$
  ::摩尔质量=1.211 g0.0275 mol=44.0 mg/mol

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  Since N has a molar mass of 14 g/mol and O has a molar mass of 16 g/mol, the formula N ₂ O would produce the correct molar mass.
  ::由于N的摩尔质量为14克/摩尔,O的摩尔质量为16克/摩尔,N2O的配方将产生正确的摩尔质量。

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

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  The  $\begin{align*}R\end{align*}$ value that corresponds to a pressure in atm was chosen for this problem. The calculated molar mass gives a reasonable formula for dinitrogen monoxide.
  ::为这一问题选择了与原子压力相对应的R值。计算出的摩尔质量为一氧化二氮提供了一个合理的公式。

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  Calculating Density of a Gas
  ::计算气体密度

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  The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia gas (NH ₃ ) at 0.913 atm and 20°C, assuming the ammonia is ideal. First, the molar mass of ammonia is calculated to be 17.04 g/mol. Next, assume exactly 1 mol of ammonia  $\begin{align*}(n = 1)\end{align*}$ and calculate the volume that such an amount would occupy at the given temperature and pressure.
  ::理想的气体法可以用来在不标准的条件下找到气体的密度。例如,我们将确定氨气(NH3)的密度,在0.913 atm和20°C时,假设氨是理想的。首先,氨的摩尔质量计算为17.04克/摩尔。接着,精确假定氨1毫升(n=1),并计算在给定温度和压力下这一数量所占用的数量。

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  $$\begin{align*}V = \frac{ nRT}{P}=\frac{1.00 \ \text{mol} \times 0.08206 \text{ L} \cdot \text{atm/K}\cdot \text{mol} \times 293 \text{ K}}{0.913 \ \text{atm}}=26.3 \text{ L}\end{align*}$$
  ::V=nRTP=1.00 molx0.008206 Latm/Kmol×293 K0.913 atm=26.3 L

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  Now the density can be calculated by dividing the mass of one mole of ammonia by the volume above.
  ::现在,密度可以通过将氨的一个摩尔质量除以上面的体积来计算。

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  $$\begin{align*}\text{density}=\frac{17.04 \text{ g}}{26.3 \text{ L}}=0.647 \ \text{g/L}\end{align*}$$
  ::密度=17.04 g26.3 L=0.647 g/L

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  As a point of comparison, this density is slightly less than the density of ammonia at STP, which is equal to $\begin{align*}\frac{(17.04 \ \text{g/mol})}{(22.4 \ \text{L/mol})} = 0.761\ \text{g/L}\end{align*}$ . It makes sense that the density should be lower compared to that at STP since both the increase in temperature (from 0°C to 20°C) and the decrease in pressure (from 1 atm to 0.913 atm) would cause the NH ₃ molecules to spread out a bit further from one another.
  ::相比之下,这种密度略低于在STP的氨的密度,即等于(17.04克/摩尔)(22.4升/摩尔)=0.761克/升。 与STP的密度相比,这种密度应当较低,因为温度上升(从0°C到20°C)和压力下降(从1吨到0.913吨)将使NH3分子进一步扩散。

   

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   Summary
  ::摘要

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  • Calculations of molar mass and density of an ideal gas are described.
    ::描述理想气体的摩尔质量和密度的计算。

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  Review
  ::回顾

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  1. Why do you need the volume, temperature, and pressure of the gas to calculate molar mass?
     ::为什么你需要气体的体积、温度和压力来计算摩尔质量?
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  2. What assumption about the gas is made in all these calculations?
     ::在所有这些计算中,关于气体的假设是什么?
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  3. Why do you need the mass of the gas to calculate the molar mass?
     ::为什么你需要气体质量来计算摩尔质量?