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9.3: Stoichiometry of Gaseous Substances, Mixtures, and Reactions

  • Page ID
    1870
  • Skills to Develop

    • Use the ideal gas law to compute gas densities and molar masses
    • Perform stoichiometric calculations involving gaseous substances
    • State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures

    The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it."

    As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.

    Density of a Gas

    Recall that the density of a gas is its mass to volume ratio, \(ρ=\dfrac{m}{V}\). Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, PV = nRT, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example \(\PageIndex{1}\).

    Example \(\PageIndex{1}\): Derivation of a Density Formula from the Ideal Gas Law

    Use PV = nRT to derive a formula for the density of gas in g/L

    Solution

    \[PV = nRT\]

    Rearrange to get (mol/L):

    \[\dfrac{n}{v}=\dfrac{P}{RT}\]

    Multiply each side of the equation by the molar mass, ℳ. When moles are multiplied by ℳ in g/mol, g are obtained:

    \[(ℳ)\left(\dfrac{n}{V}\right)=\left(\dfrac{P}{RT}\right)(ℳ)\]

    \[ℳ/V=ρ=\dfrac{Pℳ}{RT}\]

    Exercise \(\PageIndex{1}\)

    A gas was found to have a density of 0.0847 g/L at 17.0 °C and a pressure of 760 torr. What is its molar mass? What is the gas?

    Answer

    \[ρ=\dfrac{Pℳ}{RT} \nonumber\]

    \[\mathrm{0.0847\:g/L=760\cancel{torr}×\dfrac{1\cancel{atm}}{760\cancel{torr}}×\dfrac{\mathit{ℳ}}{0.0821\: L\cancel{atm}/mol\: K}×290\: K}\]

    ℳ = 2.02 g/mol; therefore, the gas must be hydrogen (H2, 2.02 g/mol)

    We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at STP.

    Example \(\PageIndex{2}\): Empirical/Molecular Formula Problems

    Using the Ideal Gas Law and Density of a Gas Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?

    Solution

    Strategy:

    First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:

    \[\mathrm{85.7\: g\: C×\dfrac{1\: mol\: C}{12.01\: g\: C}=7.136\: mol\: C\hspace{20px}\dfrac{7.136}{7.136}=1.00\: mol\: C}\]

    \[\mathrm{14.3\: g\: H×\dfrac{1\: mol\: H}{1.01\: g\: H}=14.158\: mol\: H\hspace{20px}\dfrac{14.158}{7.136}=1.98\: mol\: H}\]

    Empirical formula is CH2 [empirical mass (EM) of 14.03 g/empirical unit].

    Next, use the density equation related to the ideal gas law to determine the molar mass:

    \[d=\dfrac{Pℳ}{RT}\hspace{20px}\mathrm{\dfrac{1.56\: g}{1.00\: L}=0.984\: atm×\dfrac{ℳ}{0.0821\: L\: atm/mol\: K}×323\: K}\]

    ℳ = 42.0 g/mol, \(\dfrac{ℳ}{Eℳ}=\dfrac{42.0}{14.03}=2.99\), so (3)(CH2) = C3H6 (molecular formula)

    Exercise \(\PageIndex{2}\)

    Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?

    Answer

    Empirical formula, CH; Molecular formula, C2H2

    Molar Mass of a Gas

    Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n:

    \[ℳ=\mathrm{\dfrac{grams\: of\: substance}{moles\: of\: substance}}=\dfrac{m}{n}\]

    The ideal gas equation can be rearranged to isolate n:

    \[n=\dfrac{PV}{RT}\]

    and then combined with the molar mass equation to yield:

    \[ℳ=\dfrac{mRT}{PV}\]

    This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.

    Example \(\PageIndex{3}\): Determining the Molar Mass of a Volatile Liquid

    The approximate molar mass of a volatile liquid can be determined by:

    1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
    2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
    3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (Figure \(\PageIndex{1}\))

    This figure shows four photos each connected by a right-facing arrow. The first photo shows a glass flask with aluminum foil covering the top sitting on a scale. The scale reads 89.516. The second photo shows a syringe being inserted into the flask through the aluminum foil covering. The third photo shows the glass flask being inserted into a beaker of water. The water appears to be heated at 100. The fourth photo shows the glass flask being weighed again. This time the scale reads 89.512. 

    Figure \(\PageIndex{1}\): When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At \(t_{l⟶g}\), the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)

    Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?

    Solution

    Since

    \[ℳ=\dfrac{m}{n} \nonumber \]

    and

    \[n=\dfrac{PV}{RT} \nonumber\]

    substituting and rearranging gives

    \[ℳ=\dfrac{mRT }{PV}\]

    then

    \[ℳ=\dfrac{mRT}{PV}=\mathrm{\dfrac{(0.494\: g)×0.08206\: L⋅atm/mol\: K×372.8\: K}{0.976\: atm×0.129\: L}=120\:g/mol} \]

    Exercise \(\PageIndex{3}\)

    A sample of phosphorus that weighs 3.243 × 10−2 g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor?

    Answer

    124 g/mol P4

    The Pressure of a Mixture of Gases: Dalton’s Law

    Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container ( Figure \(\PageIndex{2}\)). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:

    \[P_{Total}=P_A+P_B+P_C+...=\sum_iP_i\]

    In the equation PTotal is the total pressure of a mixture of gases, PA is the partial pressure of gas A; PB is the partial pressure of gas B; PC is the partial pressure of gas C; and so on.

    This figure includes images of four gas-filled cylinders or tanks. Each has a valve at the top. The interior of the first cylinder is shaded blue. This region contains 5 small blue circles that are evenly distributed. The label “300 k P a” is on the cylinder. The second cylinder is shaded lavender. This region contains 8 small purple circles that are evenly distributed. The label “600 k P a” is on the cylinder. To the right of these cylinders is a third cylinder. Its interior is shaded pale yellow. This region contains 12 small yellow circles that are evenly distributed. The label “450 k P a” is on this region of the cylinder. An arrow labeled “Total pressure combined” appears to the right of these three cylinders. This arrow points to a fourth cylinder. The interior of this cylinder is shaded a pale green. It contains evenly distributed small circles in the following quantities and colors; 5 blue, 8 purple, and 12 yellow. This cylinder is labeled “1350 k P a.”

    Figure \(\PageIndex{2}\): If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size cylinder, the total pressure of the mixture is 1350 kPa.

    The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:

    \[P_A=X_A×P_{Total}\hspace{20px}\ce{where}\hspace{20px}X_A=\dfrac{n_A}{n_{Total}}\]

    where PA, XA, and nA are the partial pressure, mole fraction, and number of moles of gas A, respectively, and nTotal is the number of moles of all components in the mixture.

    Example \(\PageIndex{2}\): The Pressure of a Mixture of Gases

    A 10.0-L vessel contains 2.50 × 10−3 mol of H2, 1.00 × 10−3 mol of He, and 3.00 × 10−4 mol of Ne at 35 °C.

    1. What are the partial pressures of each of the gases?
    2. What is the total pressure in atmospheres?

    Solution

    The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using \(P=\dfrac{nRT}{V}\):

    \[P_\mathrm{H_2}=\mathrm{\dfrac{(2.50×10^{−3}\:mol)(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=6.32×10^{−3}\:atm}\]

    \[P_\ce{He}=\mathrm{\dfrac{(1.00×10^{−3}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=2.53×10^{−3}\:atm}\]

    \[P_\ce{Ne}=\mathrm{\dfrac{(3.00×10^{−4}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=7.58×10^{−4}\:atm}\]

    The total pressure is given by the sum of the partial pressures:

    \[P_\ce{T}=P_\mathrm{H_2}+P_\ce{He}+P_\ce{Ne}=\mathrm{(0.00632+0.00253+0.00076)\:atm=9.61×10^{−3}\:atm}\]

    Exercise \(\PageIndex{2}\)

    A 5.73-L flask at 25 °C contains 0.0388 mol of N2, 0.147 mol of CO, and 0.0803 mol of H2. What is the total pressure in the flask in atmospheres?

    Answer

    1.137 atm

    Here is another example of this concept, but dealing with mole fraction calculations.

    Example \(\PageIndex{3}\): The Pressure of a Mixture of Gases

    A gas mixture used for anesthesia contains 2.83 mol oxygen, O2, and 8.41 mol nitrous oxide, N2O. The total pressure of the mixture is 192 kPa.

    1. What are the mole fractions of O2 and N2O?
    2. What are the partial pressures of O2 and N2O?

    Solution

    The mole fraction is given by

    \[X_A=\dfrac{n_A}{n_{Total}} \nonumber\]

    and the partial pressure is

    \[P_A = X_A  \times P_{Total} \nonumber\]

    For O2,

    \[X_{O_2}=\dfrac{n_{O_2}}{n_{Total}}=\mathrm{\dfrac{2.83 mol}{(2.83+8.41)\:mol}=0.252} \nonumber\]

    and

    \[P_{O_2}=X_{O_2}×P_{Total}=\mathrm{0.252×192\: kPa=48.4\: kPa} \nonumber\]

    For N2O,

    \[X_{N_2O}=\dfrac{n_{N_2O}}{n_{Total}}=\mathrm{\dfrac{8.41\: mol}{(2.83+8.41)\:mol}=0.748} \nonumber\]

    and

    \[P_{N_2O}=X_{N_2O}×P_{Total}=\mathrm{(0.748)×192\: kPa = 143.6 \: kPa} \nonumber\]

    Exercise \(\PageIndex{3}\)

    What is the pressure of a mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?

    Answer

    1.87 atm

    Collection of Gases over Water

    A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure \(\PageIndex{3}\)), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.

     This figure shows a diagram of equipment used for collecting a gas over water. To the left is an Erlenmeyer flask. It is approximately two thirds full of a lavender colored liquid. Bubbles are evident in the liquid. The label “Reaction Producing Gas” appears below the flask. A line segment connects this label to the liquid in the flask. The flask has a stopper in it through which a single glass tube extends from the open region above the liquid in the flask up, through the stopper, to the right, then angles down into a pan that is nearly full of light blue water. This tube again extends right once it is well beneath the water’s surface. It then bends up into an inverted flask which is labeled “Collection Flask.” This collection flask is positioned with its mouth beneath the surface of the light blue water and appears approximately half full. Bubbles are evident in the water in the inverted flask. The open space above the water in the inverted flask is labeled “collected gas.”

    Figure \(\PageIndex{3}\): When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).

    However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor.

    A graph is shown. The horizontal axis is labeled “Temperature ( degrees C )” with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled “Vapor pressure ( torr )” with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, “Vapor pressure at ( 100 degrees C ).”

    Figure \(\PageIndex{4}\): This graph shows the vapor pressure of water at sea level as a function of temperature.

    The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure \(\PageIndex{4}\)); more detailed information on the temperature dependence of water vapor can be found in Table \(\PageIndex{1}\), and vapor pressure will be discussed in more detail in the next chapter on liquids.

    Table \(\PageIndex{1}\): Vapor Pressure of Ice and Water in Various Temperatures at Sea Level
    Temperature (°C) Pressure (torr)   Temperature (°C) Pressure (torr)   Temperature (°C) Pressure (torr)
    –10 1.95   18 15.5   30 31.8
    –5 3.0 19 16.5 35 42.2
    –2 3.9 20 17.5 40 55.3
    0 4.6 21 18.7 50 92.5
    2 5.3 22 19.8 60 149.4
    4 6.1 23 21.1 70 233.7
    6 7.0 24 22.4 80 355.1
    8 8.0 25 23.8 90 525.8
    10 9.2 26 25.2 95 633.9
    12 10.5 27 26.7 99 733.2
    14 12.0 28 28.3 100.0 760.0
    16 13.6 29 30.0 101.0 787.6
     

    Example \(\PageIndex{4}\): Pressure of a Gas Collected Over Water

    If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure \(\PageIndex{3}\), what is the partial pressure of argon?

    Solution

    According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:

    \[P_\ce{T}=P_\ce{Ar}+P_\mathrm{H_2O}\]

    Rearranging this equation to solve for the pressure of argon gives:

    \[P_\ce{Ar}=P_\ce{T}−P_\mathrm{H_2O}\]

    The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so:

    \[P_\ce{Ar}=\mathrm{750\:torr−25.2\:torr=725\:torr}\]

    Exercise \(\PageIndex{4}\)

    A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?

    Answer

    0.583 L

    Chemical Stoichiometry and Gases

    Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions. We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.

    Avogadro’s Law Revisited

    Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

    We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to

    \[\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g)\]

    a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.

    The explanation for this is illustrated in Figure \(\PageIndex{4}\). According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2.

    This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled “N subscript 2”. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled “H subscript 2.” Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled “N H subscript 3.” Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded. 

    Figure \(\PageIndex{5}\): One volume of N2 combines with three volumes of H2 to form two volumes of NH3.

    Example \(\PageIndex{5}\): Reaction of Gases

    Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.

    Solution

    The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction:

    \[\begin{align}
    &\ce{C3H8}(g)+\ce{5O2}(g) ⟶ &&\ce{3CO2}(g)+\ce{4H2O}(l)\\
    \ce{&1\: volume + 5\: volumes &&3\: volumes + 4\: volumes}
    \end{align}\]

    From the equation, we see that one volume of C3H8 will react with five volumes of O2:

    \[\mathrm{2.7\cancel{L\:C_3H_8}×\dfrac{5\: L\:\ce{O2}}{1\cancel{L\:C_3H_8}}=13.5\: L\:\ce{O2}}\]

    A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.

    Exercise \(\PageIndex{5}\)

    An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C2H2, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 103 L of O2 at 0 °C and 1 atm, will be required to burn the acetylene?

    \[\ce{2C2H2 + 5O2⟶4CO2 + 2H2O} \nonumber\]

    Answer

    3.34 tanks (2.34 × 104 L)

    Example \(\PageIndex{6}\): Volumes of Reacting Gases

    Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?

    \[\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \nonumber\]

    Solution

    Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:

    \[\mathrm{683\cancel{billion\:ft^3\:NH_3}×\dfrac{3\: billion\:ft^3\:H_2}{2\cancel{billion\:ft^3\:NH_3}}=1.02×10^3\:billion\:ft^3\:H_2}\]

    The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)

    Exercise \(\PageIndex{6}\)

    What volume of O2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C2H4(g), measured under the same conditions of temperature and pressure? The products are CO2 and water vapor.

    Answer

    51.0 L

    Example \(\PageIndex{7}\): Volume of Gaseous Product

    What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

    \[\ce{2Ga}(s)+\ce{6HCl}(aq)⟶\ce{2GaCl3}(aq)+\ce{3H2}(g)\]

    Solution

    To convert from the mass of gallium to the volume of H2(g), we need to do something like this:

    This figure shows four rectangles. The first is shaded yellow and is labeled “Mass of G a.” This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled “Moles of G a.” This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled “Moles of H subscript 2 ( g ).” This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded lavender and is labeled “Volume of H subscript 2 ( g ).”

    The first two conversions are:

    \[\mathrm{8.88\cancel{g\: Ga}×\dfrac{1\cancel{mol\: Ga}}{69.723\cancel{g\: Ga}}×\dfrac{3\: mol\:H_2}{2\cancel{mol\: Ga}}=0.191\:mol\: H_2}\]

    Finally, we can use the ideal gas law:

    \[V_\mathrm{H_2}=\left(\dfrac{nRT}{P}\right)_\mathrm{H_2}=\mathrm{\dfrac{0.191\cancel{mol}×0.08206\: L\cancel{atm\:mol^{−1}\:K^{−1}}×300\: K}{0.951\:atm}=4.94\: L}\]

    Exercise \(\PageIndex{7}\)

    Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO2 at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?

    Answer

    1.30 × 103 L

    Greenhouse Gases and Climate Change

    The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost \(\dfrac{1}{3}\) is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions—without atmosphere, the average global average temperature of 14 °C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure \(\PageIndex{6}\)).

     

    Figure \(\PageIndex{6}\): Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.

    There is strong evidence from multiple sources that higher atmospheric levels of CO2 are caused by human activity, with fossil fuel burning accounting for about \(\dfrac{3}{4}\) of the recent increase in CO2. Reliable data from ice cores reveals that CO2 concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO2 concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure \(\PageIndex{7}\)).

    This figure has the heading “Carbon Dioxide in the Atmosphere.” The first graph has a horizontal axis label “Year ( B C )” and a vertical axis label “Carbon dioxide concentration ( p p m ).” The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.

    Figure \(\PageIndex{7}\): Figure CO2 levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years.

     

    Summary

    The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.

    Key Equations

    • PTotal = PA + PB + PC + … = ƩiPi
    • PA = XA PTotal
    • \(X_A=\dfrac{n_A}{n_{Total}}\)

    Footnotes

    1. “Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, http://www-history.mcs.st-andrews.ac.../Lagrange.html

    Glossary

    Dalton’s law of partial pressures
    total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.
    mole fraction (X)
    concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components
    partial pressure
    pressure exerted by an individual gas in a mixture
    vapor pressure of water
    pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature

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