Sample Problems

4.2 Writing and Balancing Chemical Equations

A chemical reaction is a process in which one or more substances convert into one or more different substances through the breaking and forming of chemical bonds. Compounds undergo structural transformations during these processes, such as hydrogen gas combining with oxygen gas to yield water. A combustion reaction is a specific class of chemical reaction wherein a compound combines with molecular oxygen to form one or more oxygen-containing compounds while releasing significant thermal energy. Society relies heavily on the heat released from combustion reactions, such as the combustion of gasoline in automobile engine cylinders to propel vehicles or the combustion of natural gas for residential heating and cooking. Chemical reactions are represented symbolically through balanced chemical equations, where the reactant formulas appear on the left side and the product formulas appear on the right side, as illustrated by the combustion of methane:

$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(g)}$

Sample Problem: Balancing Combustion Equations

To balance the complete combustion reaction of propane gas ($C_3H_{8(g)}$) in oxygen gas ($O_{2(g)}$) producing carbon dioxide ($CO_{2(g)}$) and steam ($H_2O_{(g)}$), begin by balancing the carbon atoms, followed by hydrogen atoms, and finally oxygen atoms. Propane contains three carbon atoms and eight hydrogen atoms, which requires three molecules of $CO_2$ and four molecules of $H_2O$. Summing the product oxygen atoms gives $(3 \times 2) + (4 \times 1) = 10$ oxygen atoms, requiring five molecules of $O_2$ on the reactant side:

$C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(g)}$

Expressed in grams to demonstrate the law of conservation of mass, $44.10\,g$ of $C_3H_8$ ($1\,mol$) reacts with $160.00\,g$ of $O_2$ ($5\,mol$) to generate $132.03\,g$ of $CO_2$ ($3\,mol$) and $72.07\,g$ of $H_2O$ ($4\,mol$), confirming an identical total mass of $204.10\,g$ before and after the chemical change.

4.3 Reaction Stoichiometry: How Much Carbon Dioxide?

Reaction stoichiometry encompasses the quantitative numerical relationships between amounts of reactants and products in a balanced chemical transformation. The stoichiometric coefficients assigned to each substance in a balanced equation specify the relative amounts in moles of that substance involved in the chemical change. For instance, in the balanced reaction between cobalt(III) oxide and solid carbon:

$2Co_2O_{3(s)} + 3C_{(s)} \rightarrow 4Co_{(s)} + 3CO_{2(g)}$

Determining the quantity of carbon needed to completely react with $6\,mol$ of $Co_2O_3$ requires using the molar ratio of $3\,mol$ of $C$ to $2\,mol$ of $Co_2O_3$, indicating that $9\,mol$ of $C$ are required. Similarly, the balanced combustion equation for octane provides the stoichiometric conversion factor linking fossil fuel consumption to carbon dioxide emissions:

$2C_8H_{18(l)} + 25O_{2(g)} \rightarrow 16CO_{2(g)} + 18H_2O_{(g)}$

This stoichiometric relationship reveals that burning $2\,mol$ of liquid octane directly produces $16\,mol$ of carbon dioxide gas.

Sample Problem: Stoichiometric Mass Calculations in Grams

To calculate the mass in grams of carbon dioxide gas ($CO_2$) generated when burning $571.1\,g$ of liquid octane ($C_8H_{18}$, molar mass $114.22\,g\,mol^{-1}$):

First, convert the mass in grams of octane to moles:

$n = \frac{571.1\,g}{114.22\,g\,mol^{-1}} = 5.00\,mol\,C_8H_{18}$

Next, multiply the molar amount of octane by the stoichiometric mole ratio derived directly from the balanced equation:

$n_{CO_2} = (5.00\,mol\,C_8H_{18}) \times \left(\frac{16\,mol\,CO_2}{2\,mol\,C_8H_{18}}\right) = 40.0\,mol\,CO_2$

Finally, multiply the moles of $CO_2$ by its molar mass ($44.01\,g\,mol^{-1}$) to calculate the mass in grams:

$m = (40.0\,mol) \times (44.01\,g\,mol^{-1}) = 1760\,g\,CO_2$

4.4 Stoichiometric Relationships: Limiting Reactant, Theoretical Yield, Percent Yield, and Reactant in Excess

Stoichiometric calculations often involve reactants supplied in non-stoichiometric ratios. The limiting reactant is the reactant that is entirely consumed first, thereby placing an upper limit on the maximum possible quantity of product formed. The theoretical yield represents the calculated maximum amount of product that can form based entirely on the initial quantity of the limiting reactant. Any reactant remaining after the limiting reactant has been completely consumed is classified as the reactant in excess. This concept is modeled by pizza assembly, where four crusts, ten cups of cheese, and fifteen ounces of sauce combine according to the recipe of one crust, five ounces of sauce, and two cups of cheese per pizza. In this scenario, the sauce limits the total output to three pizzas, designating tomato sauce as the limiting reactant while cheese and crust remain in excess. When reactions are carried out experimentally, the actual yield is measured and compared to the theoretical yield to compute the percent yield, which is defined as the actual yield divided by the theoretical yield multiplied by $100\%$:

$\text{Percent Yield} = \left(\frac{\text{Theoretical Yield}}{\text{Actual Yield}}\right) \times 100\%$

Sample Problem: Percent Yield Determination

If a laboratory synthesis has a calculated theoretical yield of $25.0\,g$ of product, but an experimental isolation only produces an actual yield of $21.5\,g$, the percent yield is determined by substituting these mass values into the yield equation:

$\text{Percent Yield} = \left(\frac{21.5\,g}{25.0\,g}\right) \times 100\% = 86.0\%$

4.5 Three Examples of Chemical Reactions: Combustion, Alkali Metals, and Halogens

Chemical reactivity demonstrates distinct periodic patterns across different reaction types. Combustion reactions consistently involve a substance reacting with $O_2$ to emit heat and produce oxygenated compounds. Organic compounds containing carbon and hydrogen, or carbon, hydrogen, and oxygen, consistently yield carbon dioxide and water vapor upon complete combustion, as seen in the combustion of ethanol:

$C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(g)}$

Elemental carbon combusts to produce carbon dioxide:

$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$

Elemental hydrogen combusts to yield water:

$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}$

Beyond combustion, alkali metals represent some of the most active reducing metals in the periodic table, readily transferring electrons to form ionic salts. In contrast, halogens represent the most active nonmetals, vigorously gaining electrons from metals and other reducing agents to form halide compounds.

5.1 Molecular Gastronomy and the Spherified Cherry

Molecular gastronomy merges culinary techniques with fundamental chemical principles, such as utilizing precipitation reactions in spherification. A precipitation reaction takes place when two homogeneous solutions containing dissolved ions are combined, resulting in the formation of an insoluble solid phase called a precipitate. For example, mixing aqueous solutions of lead(II) nitrate and potassium iodide generates a brilliant yellow precipitate of lead(II) iodide. In spherification, chefs encapsulate fruit juices by precipitating a thin, flexible gel membrane around juice droplets to produce culinary items such as spherified cherries.

5.2 Solution Concentration

A solution is defined as a homogeneous mixture composed of two or more substances. The majority component of the mixture is designated as the solvent, whereas the minority component dissolved within it is the solute. In an aqueous solution, water serves as the solvent. A dilute solution contains a relatively small amount of solute relative to the solvent volume, whereas a concentrated solution contains a large proportion of solute. The standard metric for quantifying solution concentration is molarity ($M$), which expresses the number of moles of solute dissolved per cubic decimeter of solution:

$M = \frac{n}{V}$

Molarity is recorded in units of $mol\,dm^{-3}$, providing a direct conversion factor between solution volume and the molar amount of dissolved solute particles.

Sample Problem: Calculating Molarity from Mass in Grams

To find the molarity of a solution prepared by dissolving $29.22\,g$ of sodium chloride ($NaCl$, molar mass $58.44\,g\,mol^{-1}$) into enough water to yield a total solution volume of $2.00\,dm^3$:

First, convert the mass in grams of $NaCl$ into moles:

$n = \frac{29.22\,g}{58.44\,g\,mol^{-1}} = 0.500\,mol\,NaCl$

Then, apply the molarity definition:

$M = \frac{0.500\,mol}{2.00\,dm^3} = 0.250\,mol\,dm^{-3}$

5.3 Solution Stoichiometry

In aqueous reaction stoichiometry, quantities of reactants and products are typically provided in terms of solution volume and molar concentration. The volume of a solution is multiplied by its molar concentration in $mol\,dm^{-3}$ to determine the amount of solute in moles. Stoichiometric coefficients from the balanced chemical equation then convert the moles of one reactant or product into equivalent moles of another substance. Finally, the target mass or volume can be determined.

Sample Problem: Solution Precipitation Stoichiometry in Grams

To determine the mass in grams of solid lead(II) chloride ($PbCl_2$, molar mass $278.10\,g\,mol^{-1}$) precipitated when $0.150\,dm^3$ of $0.175\,mol\,dm^{-3}$ $Pb(NO_3)_{2(aq)}$ reacts completely with an excess of potassium chloride solution:

First, determine the moles of lead(II) nitrate present by multiplying volume by concentration:

$n = (0.150\,dm^3) \times (0.175\,mol\,dm^{-3}) = 0.02625\,mol\,Pb(NO_3)_2$

According to the balanced equation $Pb(NO_3)_{2(aq)} + 2KCl_{(aq)} \rightarrow PbCl_{2(s)} + 2KNO_{3(aq)}$, one mole of $Pb(NO_3)_2$ produces one mole of solid $PbCl_2$, yielding $0.02625\,mol$ of $PbCl_2$.

Convert this molar quantity to mass in grams using the molar mass of lead(II) chloride:

$m = (0.02625\,mol) \times (278.10\,g\,mol^{-1}) = 7.30\,g\,PbCl_2$

5.4 Types of Aqueous Solutions and Solubility

When a solid solute is placed into a liquid solvent, dissolution depends on a thermodynamic competition between solute–solute attractive forces holding the lattice together and solvent–solute attractive interactions. In the dissolution of sodium chloride ($NaCl$), the ionic bonds holding $Na^+$ cations and $Cl^-$ anions together compete against ion-dipole attractions between the ions and polar water molecules. Because oxygen is more electronegative than hydrogen, water possesses a permanent molecular dipole with a partial negative charge on oxygen and partial positive charges on hydrogen. Water molecules hydrate the dissociated ions, allowing ionic solids to dissolve and form electrolytic solutions. In contrast, molecular compounds like table sugar ($C_{12}H_{22}O_{11}$) dissolve without dissociating into ions, remaining intact neutral molecules surrounded by water through hydrogen bonding.

5.5 Precipitation Reactions

Precipitation reactions in aqueous environments occur when combinations of dissolved cations and anions form an insoluble ionic solid according to empirical solubility guidelines. A practical application occurs in the treatment of hard water, which contains elevated concentrations of dissolved $Ca^{2+}$ and $Mg^{2+}$ cations that diminish the cleaning efficacy of soap by forming insoluble soap scum. Laundry detergents frequently incorporate soluble sodium carbonate ($Na_2CO_{3(aq)} \rightarrow 2Na^+_{(aq)} + CO_{3(aq)}^{2-}$) to remediate hard water. The dissolved carbonate anions combine with the dissolved alkaline earth cations, precipitating insoluble calcium carbonate ($CaCO_{3(s)}$) and magnesium carbonate ($MgCO_{3(s)}$), thereby removing the problematic ions from solution.

5.6 Representing Aqueous Reactions: Molecular, Ionic, and Net Ionic Equations

Aqueous ionic reactions can be expressed through three distinct equations. A molecular equation writes neutral chemical formulas for every reactant and product as if they existed as intact compounds:

$Pb(NO_3)_{2(aq)} + 2KCl_{(aq)} \rightarrow PbCl_{2(s)} + 2KNO_{3(aq)}$

A complete ionic equation displays all soluble strong electrolytes dissociated into their individual aqueous ions, explicitly reflecting the actual species present in the mixture:

$Pb^{2+}_{(aq)} + 2NO_{3(aq)}^- + 2K^+_{(aq)} + 2Cl^-_{(aq)} \rightarrow PbCl_{2(s)} + 2K^+_{(aq)} + 2NO_{3(aq)}^-$

In this complete ionic representation, $K^+_{(aq)}$ and $NO_{3(aq)}^-$ appear unaltered on both sides of the reaction arrow. These species do not participate in the chemical transformation and are termed spectator ions. Canceling all spectator ions yields the net ionic equation, which depicts exclusively the species that undergo chemical change:

$Pb^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow PbCl_{2(s)}$

5.7 Acid–Base Reactions

An acid–base reaction, also known as a neutralization reaction, occurs when an acid reacts with a base to produce water and an aqueous ionic salt, neutralizing the acidic and basic properties of the reactants. An Arrhenius acid produces $H^+$ ions upon dissolution in water, whereas an Arrhenius base produces $OH^-$ ions. In human physiology, stomach acid consists of hydrochloric acid ($HCl$), which aids digestion but causes heartburn when present in excess. Antacids alleviate acidity through neutralization reactions using bases such as magnesium hydroxide ($Mg(OH)_2$) or aluminum hydroxide ($Al(OH)_3$) to react with hydronium ions and generate neutral water.

5.8 Gas-Evolution Reactions

Gas-evolution reactions occur when the combination of aqueous reactants results in a gaseous product that bubbles out of the mixture. Some reactions evolve gas directly via ion exchange, such as sulfuric acid reacting with lithium sulfide to generate dihydrogen sulfide gas directly:

$H_2SO_{4(aq)} + Li_2S_{(aq)} \rightarrow H_2S_{(g)} + Li_2SO_{4(aq)}$

Other gas-evolution reactions proceed through an unstable intermediate that spontaneously decomposes into a gas and liquid water. Mixing hydrochloric acid with sodium bicarbonate initially produces carbonic acid ($H_2CO_3$), which decomposes rapidly into gaseous carbon dioxide and liquid water:

$HCl_{(aq)} + NaHCO_{3(aq)} \rightarrow H_2O_{(l)} + CO_{2(g)} + NaCl_{(aq)}$

5.9 Oxidation–Reduction Reactions

Oxidation–reduction reactions, commonly referred to as redox reactions, are chemical processes that involve the transfer of electrons from one reactant species to another. Classic examples include corrosion, such as the rusting of iron:

$4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)}$

The combustion of hydrocarbons:

$2C_8H_{18(l)} + 25O_{2(g)} \rightarrow 16CO_{2(g)} + 18H_2O_{(g)}$

And the synthesis of water from molecular hydrogen and oxygen:

$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}$

Redox reactions do not require elemental oxygen; any electron transfer qualifies, such as the direct synthesis of sodium chloride from elemental sodium and chlorine gas:

$2Na_{(s)} + Cl_{2(g)} \rightarrow 2NaCl_{(s)}$

In this reaction, neutral sodium atoms lose electrons to become oxidized to $Na^+$ cations, while elemental chlorine molecules gain electrons to become reduced to $Cl^-$ anions.

6.1 Supersonic Skydiving and the Risk of Decompression

Pressure ($P$) is defined as the force ($F$) exerted per unit surface area ($A$):

$P = \frac{F}{A}$

Gas pressure originates from the billions of continuous collisions that gas atoms and molecules make against surrounding surfaces. The magnitude of gas pressure depends directly on the concentration or density of particles within the volume; lower particle concentrations generate fewer collisions and therefore lower pressure. At extreme altitudes, such as $41.4\,km$ above Earth's surface where Alan Eustace completed his record supersonic skydive, the atmospheric particle density is exceedingly low. Without a specialized pressurized suit providing an artificial atmosphere near terrestrial surface pressure, human survival is impossible due to decompression and severe hypoxia.

Sample Problem: Pressure from Mass in Grams and Area

To compute the pressure exerted when an object having a mass of $51000\,g$ ($51.0\,kg$, exerting a downward gravitational force of $500\,N$ via $F = m \times g$) rests uniformly over a surface area of $0.250\,m^2$, apply the formula for pressure:

$P = \frac{500\,N}{0.250\,m^2} = 2000\,Pa$

6.2 Pressure: The Result of Molecular Collisions

The macroscopic force exerted by a gas is the collective result of countless submicroscopic particle collisions over a given area. Atmospheric pressure governs essential terrestrial phenomena, enabling humans to breathe, drink through straws, and sustain weather systems driven by pressure gradients. Because the gravitational pull on gas particles causes gas density to decrease with increasing elevation, atmospheric pressure decreases monotonically with increasing altitude. At cruising altitudes above $30000\,ft$, ambient atmospheric pressure is insufficient to maintain adequate blood oxygenation, requiring commercial aircraft cabins to remain artificially pressurized.

6.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law

The physical state of any gas sample is characterized by four state variables: pressure ($P$), volume ($V$), absolute temperature ($T$), and amount in moles ($n$). The simple gas laws define empirical relationships between pairs of these variables while holding the remaining two variables constant. Boyle’s law dictates that the volume of a gas is inversely proportional to its pressure at constant temperature and molar amount ($V \propto \frac{1}{P}$, or $P_1V_1 = P_2V_2$). Charles’s law establishes that the volume of a gas is directly proportional to its absolute temperature in Kelvin when pressure and molar amount remain constant ($V \propto T$, or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$). Avogadro’s law states that the volume of a gas is directly proportional to the amount of gas in moles at constant temperature and pressure ($V \propto n$, or $\frac{V_1}{n_1} = \frac{V_2}{n_2}$).

Sample Problem: Boyle’s and Charles’s Laws

For Boyle's law, consider a $5.60\,g$ sample of nitrogen gas ($N_2$, molar mass $28.02\,g\,mol^{-1}$) occupying $4.00\,dm^3$ at $1.00\,atm$. When compressed isothermally to a pressure of $2.50\,atm$, the new volume is determined by rearranging $P_1V_1 = P_2V_2$:

$V_2 = \frac{(1.00\,atm) \times (4.00\,dm^3)}{2.50\,atm} = 1.60\,dm^3$

For Charles's law, the same $5.60\,g$ sample occupying $2.00\,dm^3$ at an initial temperature of $300\,K$ expands upon heating isobarically to $450\,K$. The final volume is obtained by rearranging $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:

$V_2 = \frac{(2.00\,dm^3) \times (450\,K)}{300\,K} = 3.00\,dm^3$

6.4 The Ideal Gas Law

Combining Boyle’s, Charles’s, and Avogadro’s laws yields a single unified proportionality stating that gas volume is proportional to molar amount and temperature, and inversely proportional to pressure ($V \propto \frac{nT}{P}$). Introducing the ideal gas constant ($R$) as the proportionality constant produces the ideal gas law:

$PV = nRT$

An ideal gas represents a hypothetical model that follows this equation under all thermodynamic conditions. The universal gas constant $R$ has a value of $0.08206\,atm\,dm^3\,mol^{-1}\,K^{-1}$. The ideal gas law inherently contains each of the individual simple gas laws, allowing direct algebraic calculation of any single gas variable when the other three parameters are specified.

Sample Problem: Ideal Gas Law Calculation from Mass in Grams

To calculate the pressure exerted by $16.00\,g$ of oxygen gas ($O_2$, molar mass $32.00\,g\,mol^{-1}$) confined within a container of volume $10.0\,dm^3$ at a temperature of $300\,K$:

First, convert the mass in grams of $O_2$ into moles:

$n = \frac{16.00\,g}{32.00\,g\,mol^{-1}} = 0.500\,mol\,O_2$

Then, isolate pressure in the ideal gas equation:

$P = \frac{nRT}{V} = \frac{(0.500\,mol) \times (0.08206\,atm\,dm^3\,mol^{-1}\,K^{-1}) \times (300\,K)}{10.0\,dm^3} = 1.23\,atm$

6.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas

The molar volume of a gas is the volume occupied by one mole of substance. Under standard temperature and pressure (STP, defined as $T = 273\,K$ and $P = 1.00\,atm$), substituting these conditions into the ideal gas law reveals that one mole of any ideal gas occupies approximately $22.4\,dm^3$:

$V = \frac{(1.00\,mol) \times (0.08206\,atm\,dm^3\,mol^{-1}\,K^{-1}) \times (273\,K)}{1.00\,atm} = 22.4\,dm^3$

The ideal gas law is also applied to determine gas density by substituting molar mass ($M$) and sample mass in grams ($m$) into the equation, yielding:

$\rho = \frac{P \times M}{R \times T}$

Rearranging this relationship enables the experimental determination of the molar mass of an unknown volatile compound from measurements of sample mass, pressure, temperature, and volume.

Sample Problem: Calculating Gas Density in Grams per Volume

To find the density in grams per cubic decimeter of diatomic oxygen gas ($O_2$, molar mass $32.00\,g\,mol^{-1}$) under standard temperature and pressure ($P = 1.00\,atm$ and $T = 273\,K$), evaluate the density expression:

$\rho = \frac{(1.00\,atm) \times (32.00\,g\,mol^{-1})}{(0.08206\,atm\,dm^3\,mol^{-1}\,K^{-1}) \times (273\,K)} = 1.43\,g\,dm^{-3}$

This confirms that each $1.00\,dm^3$ of oxygen gas at STP has an equivalent mass of $1.43\,g$.

6.6 Mixtures of Gases and Partial Pressures

Most practical gas samples exist as homogeneous mixtures, such as dry air containing nitrogen, oxygen, argon, and carbon dioxide. Because particles in an ideal gas behave independently without mutual intermolecular interactions, each component gas within a mixture exerts a partial pressure equal to the pressure it would exert if it occupied the entire container volume alone. According to Dalton’s law of partial pressures, the total pressure of a gas mixture equals the sum of the partial pressures of all component gases:

$P_{\text{total}} = P_1 + P_2 + P_3 + \dots$

In ambient dry air, nitrogen gas accounts for approximately $78\%$ of the total pressure and oxygen gas accounts for approximately $21\%$, with each gas behaving entirely independently of the other components.

Sample Problem: Dalton’s Law of Partial Pressures

A container holds a gaseous mixture consisting of nitrogen gas with a partial pressure of $0.780\,atm$, oxygen gas with a partial pressure of $0.210\,atm$, and argon gas with a partial pressure of $0.009\,atm$. The total pressure exerted by the gas mixture is found by summing the partial pressures:

$P_{\text{total}} = 0.780\,atm + 0.210\,atm + 0.009\,atm = 0.999\,atm$

6.7 Gases in Chemical Reactions: Stoichiometry Revisited

Stoichiometric calculations involving gaseous reactants or products link measured gas volumes directly to molar amounts using the ideal gas law. Rather than relying solely on mass measurements, the moles of a gas are calculated from known pressure, volume, and temperature using:

$n = \frac{PV}{RT}$

Once the molar amount is calculated, balanced equation coefficients provide the conversion factors to determine the theoretical yield of products or the quantity of reacting partners required. Conversely, stoichiometric mole values derived from solid or solution reactants convert into expected gas volumes at specified reaction conditions using:

$V = \frac{nRT}{P}$

Sample Problem: Gas Stoichiometry Calculation from Mass in Grams

To compute the volume and mass in grams of carbon dioxide gas evolved at $1.00\,atm$ and $298\,K$ from the complete thermal decomposition of $10.01\,g$ of calcium carbonate ($CaCO_3$, molar mass $100.09\,g\,mol^{-1}$) via $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$:

First, convert the mass in grams of $CaCO_3$ into moles:

$n = \frac{10.01\,g}{100.09\,g\,mol^{-1}} = 0.100\,mol\,CaCO_3$

Because the balanced equation produces one mole of $CO_2$ per mole of $CaCO_3$, $0.100\,mol$ of $CO_2$ is produced.

Substitute this into the ideal gas volume formula:

$V = \frac{(0.100\,mol) \times (0.08206\,atm\,dm^3\,mol^{-1}\,K^{-1}) \times (298\,K)}{1.00\,atm} = 2.45\,dm^3$

Finally, multiply the moles of evolved $CO_2$ by its molar mass ($44.01\,g\,mol^{-1}$) to calculate the mass of gas in grams:

$m = (0.100\,mol) \times (44.01\,g\,mol^{-1}) = 4.40\,g\,CO_2$

6.8 Kinetic Molecular Theory: A Model for Gases

Kinetic molecular theory provides a mechanical model explaining why gases behave in accordance with macroscopic gas laws. The theory treats gases as collections of individual particles in continuous, random, straight-line motion until colliding with container boundaries or other particles. This theoretical model rests upon three core postulates that govern gas behavior.

First, the volume occupied by individual gas particles is negligibly small compared to the vast total volume of empty space separating them, justifying treating particle volume as zero. Second, the average kinetic energy of gas particles is directly proportional to absolute temperature in Kelvin, meaning all gases at identical temperatures share identical average kinetic energies regardless of their chemical identity or molar mass. Third, collisions between gas particles, or between particles and container surfaces, are completely elastic, which means total kinetic energy is conserved without attractive or repulsive intermolecular forces acting between particles.

6.9 Mean Free Path, Diffusion, and Effusion of Gases

Although gaseous molecules move at root mean square speeds of hundreds of meters per second at room temperature, net gaseous transport across space is slowed by molecular collisions. Gas molecules experience billions of collisions per second, resulting in a random, haphazard path. The mean free path defines the average distance a particle travels between successive collisions. At room temperature and atmospheric pressure, nitrogen gas has a mean free path of approximately $93\,nm$, which corresponds to roughly $310$ times its molecular diameter. This rapid succession of collisions explains why gaseous diffusion—the process by which gas molecules spread through space in response to a concentration gradient—requires minutes for a scent to cross a room rather than occurring instantaneously.

6.10 Real Gases: The Effects of Size and Intermolecular Forces

Real gases deviate from ideal behavior under extreme conditions, specifically at elevated pressures and reduced temperatures where kinetic molecular theory assumptions break down. At STP, the molar volumes of most real gases are very close to the ideal value of $22.41\,dm^3$. However, under high external pressure, the volume of the container becomes small enough that the physical volume occupied by the particles themselves represents a substantial fraction of the total space, causing the actual molar volume to exceed ideal predictions. Under low temperatures, particles possess reduced kinetic energies and move more slowly, allowing attractive intermolecular forces to pull neighboring molecules together during close passes. These attractive forces cushion impacts against container walls, reducing the observed macroscopic pressure relative to the ideal gas law prediction.