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  • 卷体口测量

    A propane grill with an open lid, positioned next to a propane tank in a backyard.

    How much propane is left in the tank?
    ::坦克里还剩多少丙烷?

    As the weather gets warmer, more and more people want to cook out on the back deck or back yard. Many folks still use charcoal for grilling because of the added flavor. But increasing numbers of back yard cooks like to use a propane grill. The burns clean, the grill is ready to go as soon as the flame is lit – but how do you know how much propane is left in the tank? You can buy gauges at hardware stores that measure and tell you how much is left in the tank.
    ::随着天气变暖,越来越多的人想在后甲板或后院做饭。 许多人仍然使用木炭做烧烤,因为添加的口味。 但越来越多的后院厨师喜欢使用丙烷烧烤。 烧烤干净,烧烤准备就绪,一旦火焰燃起 — — 但你知道罐中还剩下多少丙烷吗?你可以在计量并告诉你罐中剩下多少的硬体商店购买测量仪。

    Volume-Volume Stoichiometry
    ::卷体口测量

    Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain the same number of gas particles. Further, one mole of any gas at standard temperature and pressure (0°C and 1 atm) occupies a volume of 22.4 L. These characteristics make stoichiometry problems involving gases at STP very straightforward. Consider the reaction of nitrogen and oxygen cases to form nitrogen dioxide.
    ::Avogadro的假设指出,在相同的温度和压力下,所有气体的同等数量都含有相同数量的气体颗粒,此外,标准温度和压力(0°C和1 atm)下的任何气体的一毫尔(0°C和1 atm)占22.4升的体积。 这些特性使得在STTP上涉及气体的声学测量问题非常简单。 想想氮和氧气体反应形成二氧化氮。

    N 2 ( g )     +   2 O 2 ( g ) → 2 NO 2 ( g ) 1   molecule 2   molecules 2   molecules 1   mol   2   mol   2   mol 1   volume     2   volumes     2   volumes

    ::N2(g) + 2O2(g) + 2O2(g) 2NO2(g)1分子2分子2分子2分子1 mol 2 mol 2 mol1 第2卷第2卷第2卷

    Because of Avogadro’s work, we know that the between substances in a gas- phase reaction are also volume ratios. The six possible volume ratios for the above equation are:
    ::由于Avogadro的工作,我们知道,气相反应物质之间的体积比率也是体积比率。

    1. 1   volume N 2 2   volumes O 2     o r   2   volumes O 2 1   volume N 2
      ::1卷N22 卷O2或2卷O2 卷O21 卷N2
    2. 1   volume N 2 2   volumes NO 2   o r   2   volumes NO 2 1   volume N 2
      ::1卷N22卷NO2或2卷NO21卷N2
    3. 2   volumes O 2 2   volumes NO 2   o r   2   volumes NO 2 2   volumes O 2
      ::2卷O22卷NO2或2卷NO22卷O2

    The volume ratios above can easily be used when the volume of one gas in a reaction is known and you need to determine the volume of another gas that will either react with or be produced from the first gas. The pressure and temperature conditions of both gases need to be the same.
    ::当一种气体在反应中的体积为已知时,就可以很容易地使用以上体积比率,而你需要确定另一种气体的体积,这种气体要么与第一种气体发生反应,要么与第一种气体发生反应。两种气体的压力和温度条件必须相同。

    Sample Problem: Volume-Volume Stoichiometry
    ::抽样问题: 量-量 - 量 - 量 - 量 - 量 - 量 - 量 - 度

    The combustion of propane gas produces carbon dioxide and water vapor .
    ::丙烷气体的燃烧产生二氧化碳和水蒸气。

    C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 CO 2 ( g ) + 4 H 2 O ( g )

    ::C3H8(g)+5O2(g)+5O2(g)+3CO2(g)+4H2O(g)

    What volume of oxygen is required to completely combust 0.650 L of propane? What volume of carbon dioxide is produced in the reaction?
    ::完全燃烧0.650升丙烷需要多少氧气?反应产生的二氧化碳量是多少?

    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: 0.650 L C 3 H 8
      ::说明:0.650 L C3H8
    • 1 volume C 3 H 8 = 5 volumes O 2
      ::1卷C3H8 = 5卷O2
    • 1 volume C 3 H 8 = 3 volumes CO 2
      ::1卷C3H8 = 3 卷CO2

    Unknown
    ::未知

    • volume O 2 = ? L
      ::体积O2=L
    • volume CO 2 = ? L
      ::CO2 = L 体积CO2 = L

    Two separate calculations can be done using the volume ratios.
    ::可以用量比率进行两个独立的计算。

    Step 2: Solve.
    ::步骤2:解决。

    0.650  L C 3 H 8 × 5  L O 2 1  L C 3 H 8 = 3.25  L O 2 0.650  L C 3 H 8 × 3  L CO 2 1  L C 3 H 8 = 1.95  L CO 2

    ::0.650 L C3H8×5 L O21 L C3H8=3.25 L O20.650 L C3H8×3 L CO21 L C3H8=1.95 L CO2

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

    Because the coefficients of the O 2 and the CO 2 are larger than that of the C 3 H 8 , the volumes for those two gases are greater. Note that total volume is not necessarily conserved in a reaction because moles are not necessarily conserved. In this reaction, 6 total volumes of reactants become 7 total volumes of products.
    ::由于O2和CO2的系数大于C3H8的系数,这两种气体的体积更大。 请注意,总体积不一定在反应中被保存,因为摩尔不一定被保存。 在这一反应中,6个反应体的总体积成为7个产品的总体积。

    Summary
    ::摘要

    • Calculations of volume-volume ratios are based on Avogadro’s hypothesis.
      ::量-量比率的计算基于Avogadro的假设。
    • Pressures and temperatures of the gases involved need to be the same.
      ::所涉气体的压力和温度必须相同。

    Review
    ::回顾

    1. What is Avogadro’s hypothesis?
      ::Avogadro的假设是什么?
    2. How much volume is occupied by one mole of a gas at STP?
      ::受威胁受威胁受威胁人民协会一个煤气的内鬼占用了多少量?
    3. In the sample problem above, assume we combust 1.3 L of propane. How much CO 2 will be produced?
      ::在上述抽样问题中,假设我们燃烧了1.3升丙烷,将产生多少二氧化碳?