Chapter 8: Quantities in Chemical Reactions – Stoichiometry, Limiting Reactants, and Yield

Quantities in Chemical Reactions

Introduction

This chapter explores the quantitative relationships in chemical reactions, focusing on stoichiometry, limiting reactants, theoretical and actual yields, and percent yield. These concepts are essential for predicting the amounts of substances consumed and produced in chemical processes.

Global Warming and Greenhouse Gases

Combustion of Fossil Fuels

• The combustion of fossil fuels, such as octane (C8H18), produces water and carbon dioxide as products.

• Carbon dioxide (CO2) is a major greenhouse gas implicated in global warming.

The Greenhouse Effect

• Greenhouse gases allow sunlight to enter the atmosphere but prevent heat from escaping, similar to glass in a greenhouse.

• Increased greenhouse gas concentrations trap more heat, raising Earth's temperature.

• Since 1880, atmospheric CO2 levels have risen by 38%, and Earth's average temperature has increased by about 1.9°F.

Stoichiometry: Relationships Between Ingredients

Definition and Importance

• Stoichiometry is the numerical relationship between chemical quantities in a balanced chemical equation.

• It allows prediction of product amounts from given reactant quantities and vice versa.

• It also enables calculation of the amount of one reactant needed to completely react with another.

Analogy: Making Pancakes

• A recipe provides numerical relationships between ingredients and the number of pancakes produced.

• For example: 2 eggs make 5 pancakes, so 8 eggs make 20 pancakes if other ingredients are sufficient.

• This ratio concept is directly analogous to stoichiometric ratios in chemical equations.

Stoichiometry: Mole-to-Mole Conversions

Balanced Chemical Equations as Recipes

• Example: $3H_2(g) + N_2(g) \rightarrow 2NH_3(g)$

• This equation shows that 3 moles of hydrogen react with 1 mole of nitrogen to produce 2 moles of ammonia.

• Stoichiometric ratios: 3 mol H2 : 1 mol N2 : 2 mol NH3

Example Calculation

• If you have 3 mol N2 and excess H2, how much NH3 can be made?

$3\ \text{mol}\ N_2 \times \frac{2\ \text{mol}\ NH_3}{1\ \text{mol}\ N_2} = 6\ \text{mol}\ NH_3$

Stoichiometry: Mass-to-Mass Conversions

General Approach

• Convert mass of reactant (A) to moles using molar mass.

• Use the balanced equation to convert moles of A to moles of product (B).

• Convert moles of B to mass using molar mass.

Example Calculation

• Combustion of octane: $2C_8H_{18}(l) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g)$

• Given: 5.0 × 102 g octane. Find mass of CO2 produced.

$5.0 \times 10^2\ \text{g}\ C_8H_{18} \times \frac{1\ \text{mol}\ C_8H_{18}}{114.3\ \text{g}\ C_8H_{18}} \times \frac{16\ \text{mol}\ CO_2}{2\ \text{mol}\ C_8H_{18}} \times \frac{44.01\ \text{g}\ CO_2}{1\ \text{mol}\ CO_2} = 1.5 \times 10^3\ \text{g}\ CO_2$

Limiting Reactant, Theoretical Yield, and Percent Yield

Concepts and Definitions

• Limiting reactant (or limiting reagent): 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.

• Actual yield: The amount of product actually obtained from a reaction (usually less than theoretical yield).

• Percent yield: The ratio of actual yield to theoretical yield, expressed as a percentage.

$\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Analogy: Pancake Recipe

• Given: 3 cups flour, 10 eggs, 4 tsp baking powder; recipe: 1 cup flour + 2 eggs + 0.5 tsp baking powder → 5 pancakes.

• Flour allows for 15 pancakes, eggs for 25, baking powder for 40. Flour is limiting; theoretical yield is 15 pancakes.

• If only 11 pancakes are made (actual yield), percent yield is:

$\text{Percent yield} = \frac{11}{15} \times 100\% = 73\%$

Limiting Reactant Problems: Moles and Mass

Finding the Limiting Reactant (Mole Example)

• Example: $Ti(s) + 2Cl_2(g) \rightarrow TiCl_4(s)$

• Given: 1.8 mol Ti, 3.2 mol Cl2

• Calculate moles of TiCl4 from each reactant:

$1.8\ \text{mol}\ Ti \times \frac{1\ \text{mol}\ TiCl_4}{1\ \text{mol}\ Ti} = 1.8\ \text{mol}\ TiCl_4$ $3.2\ \text{mol}\ Cl_2 \times \frac{1\ \text{mol}\ TiCl_4}{2\ \text{mol}\ Cl_2} = 1.6\ \text{mol}\ TiCl_4$

• Cl2 is limiting; theoretical yield is 1.6 mol TiCl4.

Finding the Limiting Reactant (Mass Example)

• Example: $2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$

• Given: 53.2 g Na, 65.8 g Cl2

• Calculate mass of NaCl from each reactant:

$53.2\ \text{g}\ Na \times \frac{1\ \text{mol}\ Na}{22.99\ \text{g}\ Na} \times \frac{2\ \text{mol}\ NaCl}{2\ \text{mol}\ Na} \times \frac{58.44\ \text{g}\ NaCl}{1\ \text{mol}\ NaCl} = 135\ \text{g}\ NaCl$ $65.8\ \text{g}\ Cl_2 \times \frac{1\ \text{mol}\ Cl_2}{70.90\ \text{g}\ Cl_2} \times \frac{2\ \text{mol}\ NaCl}{1\ \text{mol}\ Cl_2} \times \frac{58.44\ \text{g}\ NaCl}{1\ \text{mol}\ NaCl} = 108\ \text{g}\ NaCl$

• Cl2 is limiting; theoretical yield is 108 g NaCl.

Percent Yield Example

• Given: Actual yield = 86.4 g NaCl, Theoretical yield = 108 g NaCl

$\text{Percent yield} = \frac{86.4\ \text{g}}{108\ \text{g}} \times 100\% = 80.0\%$

Summary Table: Key Terms and Definitions

Term	Definition
Stoichiometry	Numerical relationship between reactants and products in a balanced chemical equation
Limiting Reactant	Reactant that is completely consumed, limiting the amount of product formed
Theoretical Yield	Maximum amount of product possible from the limiting reactant
Actual Yield	Amount of product actually obtained from a reaction
Percent Yield	Actual yield divided by theoretical yield, multiplied by 100%

Learning Objectives

• Recognize numerical relationships in balanced chemical equations.

• Carry out mole-to-mole and mass-to-mass conversions.

• Calculate limiting reactant, theoretical yield, and percent yield.