Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams,...

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Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University

Transcript of Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams,...

Page 1: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Chapter 4: Chemical Quantities & Aqueous Reactions

CHE 123: General Chemistry I

Dr. Jerome Williams, Ph.D.

Saint Leo University

Page 2: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Overview

• Limiting Reactants

• Solutions: An Introduction

• Solution Stoichiometry

Page 3: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Limiting Reactant

• Limiting Reactant (Reagent) is defined as the reactant that is consumed first in a chemical reaction and, therefore, limits the amount of products that form.

Page 4: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Limiting Reactant

• A Practical Example

• Recipe

– 1 crust + 5 oz. sauce + 2 cups cheese => 1 pizza

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Limiting Reactants

• Scenario #1: Reactants are in 1:1 Ratio

– Reactant with smallest no. moles is limiting.

A + B → C

0.5 moles 0.25 moles ? moles

– The answer has to be 0.25 moles because when reactant B is used up, the reaction STOPS!

Page 16: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Limiting Reactants

• Scenario #2: Reactants are NOT in 1:1 Ratio

– Must find number of moles product using each reactant quantity and their coefficient in balanced reaction to establish which reactant is limiting.

A + 3 B → 2 C

2.50 moles 4.50 moles ? moles

– How many moles product will form? Ans. = 3.00 moles.

Page 17: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Limiting Reactants

• Scenario #2: Reactants are NOT in 1:1 Ratio

A + 3 B → 2 C 2.50 moles 4.50 moles ? moles

– 2.50 moles A x (2 moles C / 1 moles A) = 5.00 mol C– 4.50 moles B x (2 moles C / 3 moles B) = 3.00 mol C

– B is limiting reactant; it forms least amount product.– B reacts three times for every A that reacts; this occurs

despite having more moles when compared to reactant A.

Page 18: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Example Problem

• If we react 42.5 g Mg and 33.8 g O2, what is limiting reactant & theoretical yield?

• Reaction: 2 Mg + O2 ==> 2 MgO

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Example 4.3

• Starting with 86.3 g NO and 25.6 g H2, find the theoretical yield of NH3 in grams.

• Rxn: 2 NO (g) + 5 H2 (g) 2 NH3 (g) + 2 H2O (g)

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Solutions: An Introduction

• A solution is a homogeneous mixture that consists of two components.

– Solute is substance being dissolved in the solution.– Solvent is the dissolving environment.

– Most chemical reactions are carried out in liquid state or in solution.

– In biology & general chemistry, your solvent is water unless told otherwise.

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Solutions: An Introduction

• Molarity is an expression of solution concentration. It is defined as the number of moles of solute divided by the volume of a solution in liters.

– Molarity = M = moles solute / Vol. soln. in Liters

– By simple rearrangement one obtains the following equations.

– (Vol. soln. in Liters ) x (M) = moles solute

– Vol. soln. in Liters = moles solute / M

Page 28: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Solutions: An Introduction

• A dilution is a process by which water (or another solvent) is added to a solution of known concentration to achieve a new solution of lower concentration.

– Dilution does not alter number of moles of solute.

– Dilution equation

M1V1 = M2V2

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Solutions: An Introduction

• Some example problems to consider.

• What is the molarity of a solution prepared by adding 24.66 g of potassium nitrate to 250 mL of water?

– Answer = 0.975 M

• How does one prepare 250 mL of 1.50 M HCl from 12.0 M HCl ?

– Answer = Measure 31.25 mL of 12.0 M HCl and add to 218.75 mL of distilled water. Mix contents well before using.

Page 36: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Solution Stoichiometry

• Handy Dandy Five Step Method is still used.

– The only modification is now we have volumes and concentrations to consider.

– Use Molarity equations to find number of moles and then proceed as before.

Page 37: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Solution Stoichiometry

Page 38: Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University.

Solution Stoichiometry

• An example involving solutions.

– Sodium hydrogen carbonate (NaHCO3) is used as an antacid. One tablet requires 34.50 mL of 0.138M HCl solution for complete reaction. Determine the number of grams of NaHCO3 that one tablet contains:

– The balanced chemical reaction is shown below.

NaHCO3 + HCl → NaCl + H2O + CO2

• Answer: 0.400 grams NaHCO3