Ch 8: Stoichiometry, Limiting Reactants, and Enthalpy in Chemical Reactions

Quantitative Relationships in Chemical Reactions

Numerical Relationships in Balanced Chemical Equations

Balanced chemical equations provide the foundation for understanding how chemical quantities relate to each other. The coefficients in a balanced equation indicate the ratios in which reactants combine and products form.

• Stoichiometry: The numerical relationship between the amounts of reactants and products in a balanced chemical equation.

• Mole-to-mole conversions: Use the coefficients to convert between moles of different substances.

• Mass-to-mass conversions: Use molar masses and stoichiometric ratios to convert between masses of reactants and products.

• Limiting reactant: The reactant that is completely consumed first, limiting the amount of product formed.

• Theoretical yield: The maximum amount of product that can be formed from the limiting reactant.

• Percent yield: The efficiency of a reaction, calculated as $(\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%)$.

• Thermal energy in reactions: The amount of heat released or absorbed during a chemical reaction, quantified by the enthalpy change $(\Delta H_{\text{rxn}})$.

Stoichiometry and Its Applications

Stoichiometry in Everyday Life: The Pancake Analogy

Stoichiometry can be compared to following a recipe, where the proportions of ingredients determine the amount of product.

• Example: If 2 eggs make 5 pancakes, then 8 eggs make $8 \times \frac{5}{2} = 20$ pancakes.

• Conversion factors can be set up from the recipe, such as $1$cup flour : $5$ pancakes or $12$tsp baking powder : $5$ pancakes.

• To find the amount of baking powder for 3 cups of flour: $3 \text{ cups} \times \frac{12 \text{ tsp}}{1 \text{ cup}} = 36 \text{ tsp}$ baking powder.

Mole-to-Mole Conversions

Balanced chemical equations allow conversion between moles of reactants and products using stoichiometric ratios.

• Example: $3\text{H}_2(g) + \text{N}_2(g) \rightarrow 2\text{NH}_3(g)$

• Ratio: $3$mol H2 : $1$ mol N2 : $2$ mol NH3

• To find moles of NH3 from 3 mol N2: $3 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 6 \text{ mol NH}_3$

Mass-to-Mass Conversions

To convert between the mass of a reactant and the mass of a product, use the following steps:

1. Convert mass of substance A to moles using its molar mass.

2. Use the stoichiometric ratio from the balanced equation to convert moles of A to moles of B.

3. Convert moles of B to mass using its molar mass.

Example: Calculate the mass of CO2 produced from $5.0 \times 10^2$ g of octane (C8H18):

• Balanced equation: $2\text{C}_8\text{H}_{18}(l) + 25\text{O}_2(g) \rightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)$

• Calculation:

  • $5.0 \times 10^2 \text{ g C}_8\text{H}_{18} \times \frac{1 \text{ mol}}{114.3 \text{ g}} \times \frac{16 \text{ mol CO}_2}{2 \text{ mol C}_8\text{H}_{18}} \times \frac{44.01 \text{ g CO}_2}{1 \text{ mol CO}_2} = 1.5 \times 10^3 \text{ g CO}_2$

Limiting Reactant, Theoretical Yield, and Percent Yield

When reactants are not present in exact stoichiometric proportions, the limiting reactant determines the maximum amount of product that can be formed.

• Limiting reactant: The reactant that is completely consumed first.

• Theoretical yield: The amount of product formed from the limiting reactant.

• Actual yield: The measured amount of product obtained.

• Percent yield: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

Example: For the reaction $2\text{Al}(s) + 3\text{Cl}_2(g) \rightarrow 2\text{AlCl}_3(s)$, with 0.552 mol Al and 0.887 mol Cl2:

• From Al: $0.552 \text{ mol Al} \times \frac{2 \text{ mol AlCl}_3}{2 \text{ mol Al}} = 0.552 \text{ mol AlCl}_3$

• From Cl2: $0.887 \text{ mol Cl}_2 \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} = 0.591 \text{ mol AlCl}_3$

• Al is the limiting reactant; theoretical yield is 0.552 mol AlCl3.

Thermal Energy and Enthalpy in Chemical Reactions

Enthalpy of Reaction (ΔHrxn)

The enthalpy of reaction, $\Delta H_{\text{rxn}}$, measures the heat released or absorbed during a chemical reaction at constant pressure.

• Exothermic reaction: Releases heat; $\Delta H_{\text{rxn}}$ is negative.

• Endothermic reaction: Absorbs heat; $\Delta H_{\text{rxn}}$ is positive.

• Example (exothermic): $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \quad \Delta H_{\text{rxn}} = -802.3 \text{ kJ}$

• Example (endothermic): $\text{N}_2(g) + \text{O}_2(g) \rightarrow 2\text{NO}(g) \quad \Delta H_{\text{rxn}} = +182.6 \text{ kJ}$

Stoichiometry Involving Enthalpy

The amount of heat emitted or absorbed is proportional to the amount of reactant used, as indicated by the balanced equation.

• Example: $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \quad \Delta H_{\text{rxn}} = -2044 \text{ kJ}$

• To find heat released from 1.18 × 104 g C3H8:

  • $1.18 \times 10^4 \text{ g} \times \frac{1 \text{ mol}}{44.11 \text{ g}} \times \frac{-2044 \text{ kJ}}{1 \text{ mol}} = -5.47 \times 10^5 \text{ kJ}$

Tables: Key Stoichiometric Relationships

Term	Definition	Formula/Example
Limiting Reactant	Reactant that is completely consumed first, limiting product formation	Compare moles of product each reactant can produce
Theoretical Yield	Maximum possible amount of product from limiting reactant	Use stoichiometry to calculate from limiting reactant
Actual Yield	Measured amount of product obtained	Experimental value
Percent Yield	Efficiency of reaction	$\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
Enthalpy of Reaction (ΔHrxn)	Heat released or absorbed at constant pressure	Negative for exothermic, positive for endothermic

Examples and Applications

• Combustion of Octane: $2\text{C}_8\text{H}_{18}(l) + 25\text{O}_2(g) \rightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)$

  • 16 moles of CO2 are produced for every 2 moles of octane burned.

• Photosynthesis: $6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{O}_2(g)$

  • Given 58.5 g CO2, mass of glucose produced: $58.5 \text{ g} \times \frac{1 \text{ mol}}{44.01 \text{ g}} \times \frac{1 \text{ mol}}{6 \text{ mol}} \times 180.2 \text{ g} = 39.9 \text{ g}$

• Neutralization: $\text{Mg(OH)}_2(aq) + 2\text{HCl}(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{MgCl}_2(aq)$

  • 5.50 g Mg(OH)2 neutralizes 6.88 g HCl.

Summary of Key Concepts

• Balanced chemical equations provide the quantitative relationships needed for stoichiometric calculations.

• Stoichiometry allows prediction of product amounts and determination of limiting reactants.

• Mass-to-mass and mole-to-mole conversions are essential for practical chemical calculations.

• Enthalpy changes ($\Delta H_{\text{rxn}}$) quantify the thermal energy exchanged in reactions.

• Percent yield evaluates the efficiency of chemical processes.

Additional info: Some calculations and context were expanded for clarity and completeness, and all equations are provided in LaTeX format as per instructions.