Chapter 5: Stoichiometry and Dalton's Law of Partial Pressures

Stoichiometry in chemistry involves taking information from the balanced chemical equation and incorporating it into calculations.
In Chapter 5, the principles of stoichiometry established in Chapter 3 are applied specifically to gas systems.
This involves relating the amount of reactant in grams or volume to the amount of product in grams or volume by utilizing the mole as a central transition point.

To determine the amount of product from a given amount of reactant, follow these steps:
Step 1: If the reactant is given in grams, convert the mass to moles using its molar mass.
Step 2: Use the stoichiometric coefficients from the balanced chemical equation to transition from moles of reactant to moles of product.
Step 3: Convert the moles of product into the final required unit, such as grams (using molar mass) or volume (using the ideal gas equation or Avocado's law).
Stoichiometric relationships can be applied between reactant-to-product, reactant-to-reactant, or product-to-product.

Problem: Calculate the volume of oxygen in units of $dm^3$ required for the complete combustion of $7.64\,\text{dm}^3$ of acetylene ( $C_2H_2$ ) measured at the same temperature ( $T$ ) and pressure ( $P$ ).
Balanced Equation: $2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}$ .
Application of Avocado's Law:
- While coefficients generally refer to moles, Avocado's law allows these coefficients to be used directly as volumes in units of $dm^3$ when $T$ and $P$ are constant.
- From the equation, $2\,\text{dm}^3$ of acetylene requires $5\,\text{dm}^3$ of oxygen.
Calculation:
- Setting up the conversion factor to cancel acetylene: $\text{Volume of O}_2 = 7.64\,\text{dm}^3\,\text{C}_2\text{H}_2 \times \frac{5\,\text{dm}^3\,\text{O}_2}{2\,\text{dm}^3\,\text{C}_2\text{H}_2}$ .
- Result: $19.1\,\text{dm}^3\,\text{O}_2$ .

Context: Sodium azide ( $NaN_3$ ) is used in automobile airbags. The impact of a collusion triggers its decomposition, producing nitrogen gas ( $N_2$ ) that quickly inflates the bag between the driver and the windshield/dashboard.
Balanced Equation: $2\text{NaN}_3 \rightarrow 2\text{Na} + 3\text{N}_2$ .
Problem: Calculate the volume of $N_2$ generated at $80^\circ\text{C}$ and $823\,\text{mmHg}$ by the decomposition of $60\,\text{g}$ of sodium azide.
Variables Identified:
- Mass of $NaN_3 = 60\,\text{g}$ .
- Temperature ( $T$ ) = $80^\circ\text{C} + 273 = 353\,\text{K}$ .
- Pressure ( $P$ ) = $\frac{823}{760}\,\text{atm}$ .
- Required: Volume ( $V$ ) of $N_2$ .
Stoichiometric Steps to find Moles of Nitrogen ( $n$ ):
- Molar Mass of $NaN_3$ : One sodium ( $Na$ ) and three nitrogens ( $N$ ) = $65.02\,\text{g}\,\text{mol}^{-1}$ .
- Moles of $NaN_3$ : $\frac{60\,\text{g}}{65.02\,\text{g}\,\text{mol}^{-1}} = 0.923\,\text{moles of NaN}_3$ .
- Moles of $N_2$ : $\text{Moles of NaN}_3 \times \frac{3\,\text{moles N}_2}{2\,\text{moles NaN}_3} = 1.38\,\text{moles N}_2$ .
Calculating Volume via Ideal Gas Equation:
- $V = \frac{nRT}{P}$ .
- $V = \frac{(1.38\,\text{mol}) \times (0.0821\,\text{dm}^3\,\text{atm}\,\text{mol}^{-1}\,\text{K}^{-1}) \times (80 + 273\,\text{K})}{(\frac{823}{760}\,\text{atm})}$ .
- Result: $36.9\,\text{dm}^3$ .

Darthine's law (Dalton's Law) accounts for individual gases within a mixture when volume ( $V$ ) and temperature ( $T$ ) are constant.
Each gas molecule in a mixture contributes partially to the total pressure.
Ideal Gas Equation for Mixtures: Only the number of moles ( $n$ ) changes between different gases in the mix; $V$ , $T$ , and $R$ are held constant.
Total Pressure ( $P_T$ ): The sum of individual partial pressures: $P_T = P_1 + P_2 + P_3 + \dots$ .
Formula for Partial Pressure ( $P_i$ ): $P_i = X_i \times P_T$ , where $X_i$ is the mole fraction and $P_T$ is the total pressure.
Mole Fraction ( $X_i$ ): The number of moles of the gas of interest divided by the total number of moles in the container: $X_i = \frac{n_i}{n_t}$ .

Problem: A mixture contains $4.46\,\text{moles}$ of neon ( $Ne$ ), $0.74\,\text{moles}$ of argon ( $Ar$ ), and $2.15\,\text{moles}$ of xenon ( $Xe$ ). Calculate individual partial pressures if the total pressure ( $P_T$ ) is $2.008\,\text{atm}$ .
Total Moles ( $n_t$ ): $4.46 + 0.74 + 2.15 = 7.35\,\text{moles}$ .
Calculation for Neon ( $Ne$ ):
- $X_{Ne} = \frac{4.46}{7.35} = 0.607$ .
- $P_{Ne} = 0.607 \times 2.08\,\text{atm} = 1.21\,\text{atm}$ .
Calculation for Argon ( $Ar$ ):
- $X_{Ar} = \frac{0.74}{7.35} = 0.1$ .
- $P_{Ar} = 0.1 \times 2.0\,\text{atm} = 0.2\,\text{atm}$ .
Calculation for Xenon ( $Xe$ ):
- $X_{Xe} = \frac{2.15}{7.35} = 0.293$ .
- $P_{Xe} = 0.293 \times 2\,\text{atm} = 0.586\,\text{atm}$ .
Verification: The sum of partial pressures ( $1.21 + 0.2 + 0.586$ ) adds back up to approximately $2\,\text{atm}$ , confirming accuracy.

In laboratory experiments, a gas of interest is often collected over water using an ediometer. This setup displaces water as the reaction proceeds.
Because water vapor is present in the gas phase, the total pressure ( $P_T$ ) read from a barometer is the sum of the desired gas pressure and the water vapor pressure ( $P_{\text{water vapor}}$ ).
Formula: $P_{\text{gas}} = P_T - P_{\text{water vapor}}$ .
Water Vapor Pressure Constants: These are temperature-dependent values.
- At $25^\circ\text{C}$ , the vapor pressure of water is $23.76\,\text{mmHg}$ .
- At $24^\circ\text{C}$ , the vapor pressure of water is $22.4\,\text{mmHg}$ .

Problem: Oxygen gas is collected as shown in Figure 5.15 at $24^\circ\text{C}$ and a historic pressure of $762\,\text{mmHg}$ . The volume collected is $128\,\text{cm}^3$ . Calculate the mass of oxygen gas obtained ( $\text{P}_{\text{water vapor}} = 22.4\,\text{mmHg}$ ).
Step 1: Find Partial Pressure of Oxygen ( $P_{O_2}$ ):
- $P_{O_2} = 762\,\text{mmHg} - 22.4\,\text{mmHg} = 739.6\,\text{mmHg}$ .
Step 2: Convert Units:
- $P = \frac{739.6}{760}\,\text{atm}$ .
- $V = \frac{128}{1000}\,\text{dm}^3 = 0.128\,\text{dm}^3$ .
- $T = 24 + 273 = 297\,\text{K}$ .
Step 3: Solve for Moles ( $n$ ) and then Mass:
- Using $n = \frac{PV}{RT}$ .
- Molar mass of $O_2 = 2 \times 16 = 32\,\text{g}\,\text{mol}^{-1}$ .
- Calculation: $\text{Mass} = \frac{P \times V}{R \times T} \times 32\,\text{g}\,\text{mol}^{-1}$ .
- Result: $0.164\,\text{g of oxygen}$ .

Scuba diving is a real-world application of gas laws, including Darlington's (Dalton's) law of partial pressure and Boyle's law.
These principles help divers manage gas mixtures and pressure changes while underwater.

Question: How do you account for individual gas pressures in a mixture?
Answer: Use Dalton's Law of Partial Pressure. If you know the total pressure, calculate the mole fraction of the specific gas and multiply it by the total pressure. If the total pressure is unknown, use the ideal gas equation specifically for that gas's volume and temperature.
Student Inquiry Opportunity: Feel free to ask questions regarding these stoichiometric problems during lecture or recitation; practice is required to master these multi-step calculations.