Stoichiometry Study Notes

Stoichiometry

Chapter 5 Overview

  • Source: Interactive General Chemistry, 2019 Macmillan Learning

Section 5.1: Mole Calculations for Chemical Reactions

  • Mole Calculation: The calculation of the number of moles of any substance involved in a chemical reaction can be derived from the number of moles of any other substance in the reaction.

Reaction Example

  • Chemical Equation:
    2 P(s) + 3 Cl2(g) → 2 PCl3(l)

  • Interpretations:

    1. Two atoms of phosphorus react with three molecules of chlorine to produce two molecules of phosphorus trichloride.

    2. Two moles of phosphorus react with three moles of chlorine to produce two moles of phosphorus trichloride.

Example 5.1

  • Task: Write all possible reacting ratios (in moles) for the reaction 2 P(s) + 3 Cl2(g) → 2 PCl3(l)

Example 5.2

  • Task: Calculate the number of moles of aluminum that will react with 3.18 mol of oxygen to form aluminum oxide.

  • Chemical Equation:
    4 Al(s) + 3 O2(g) → 2 Al2O_3(s)

Example 5.4

  • Task: Calculate the number of moles of oxygen gas produced from the decomposition of KClO3.

  • Chemical Equation:
    2 KClO3(s) → 3 O2(g) + 2 KCl(s)

  • Given: Sample of 0.1712 mol KClO3, with 0.1146 mol decomposing.

Section 5.1 Review

  • Key Takeaways:

    • Coefficients in balanced chemical equations represent the ratio of the number of moles or formula units of all substances involved in the reaction.

    • Reacting ratios can be used as conversion factors in calculations involving any two reactants/products in reactions.

Section 5.2: Mass Calculations for Chemical Reactions

  • Objective: Use the mass of one substance to determine the masses of other substances involved in a chemical reaction.

Mass Calculations

  • Note: Reacting ratios are mole ratios, not mass ratios.

  • Conversion: Use molar mass to convert between masses and moles.

Example 5.6

  • Task: Calculate the mass of chlorine in grams produced by the electrolysis of 50.0 kg NaCl.

  • Chemical Equation:
    2 NaCl(aq) + 2 H2O(l) ightarrow 2 NaOH(aq) + Cl2(g) + H_2(g)

  • Conversion Steps:

    1. Convert 50 kg NaCl to grams: 50.0 kg NaCl X 1000 g/kg.

    2. Use the molar mass of NaCl: 1 mol NaCl/58.44 g NaCl.

    3. Convert to moles of Cl₂ using the balanced equation.

Example 5.7

  • Task: Calculate moles of SO2 needed to prepare 50.0 metric tons of liquid H2SO4.

  • Chemical Equation:
    2 SO2(g) + O2(g) + 2 H2O(l) → 2 H2SO_4(l)

Section 5.2 Review

  • Three-Step Process for Mass Calculations:

    1. Convert the mass of compound A to moles of A using the molar mass of A.

    2. Convert moles of A to moles of B using the ratio of coefficients from the balanced equation.

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

Section 5.3: Problems Involving Limiting Quantities

  • Objectives: Calculate quantities produced when specified reactants are given.

Limiting Reactant Concept

  • Definition: The limiting reactant is the one that is consumed first in the reaction, limiting the amount of product formed. Any reactant that remains unreacted is in excess.

  • Completion of Reaction: The reaction goes to completion when the limiting reactant is exhausted.

Steps in Solving Limiting Reactant Problems

  1. Identify the limiting reactant from given quantities of two or more reactants.

  2. Base all subsequent calculations on the amount of the limiting reactant.

Two Methods to Identify Limiting Reactant

  1. Calculate the amount of product that can be formed from each reactant individually; the smaller amount indicates the limiting reactant.

  2. Divide the moles of each reactant by its coefficient from the balanced equation; the smallest quotient identifies the limiting reactant.

Example 5.9

  • Task: Calculate moles of reactants/products after the reaction of 1.500 mol CH3OH with 0.500 mol O2:
    2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)

  • Initial amounts:

    • CH3OH: 1.500 mol

    • O2: 0.500 mol

  • Change due to reaction: Required calculations to fill in final amounts after the reaction.

Example 5.10

  • Task: Calculate moles of products (H3PO4 and HCl) after the reaction of 0.250 mol PCl5 with 1.50 mol H2O.

  • Chemical Equation:
    PCl5 + 4 H2O → H3PO4 + 5 HCl

Section 5.3 Review

  • Key Points:

    • The limiting reactant limits the quantity of product formed.

    • Methods to determine the limiting reactant involve calculating product yield or finding the smallest quotient of moles to coefficients.

Section 5.4: Theoretical Yield and Percent Yield

  • Objective: Express quantity of product obtained as a percentage of the maximum theoretical yield.

Definitions

  • Theoretical Yield: The maximum possible amount of product formed based on reactant quantities available.

  • Actual Yield: The amount of product actually obtained during the experimental reaction.

  • Percent Yield: The formula to calculate percent yield is:
    ext{percent yield} = rac{ ext{actual yield}}{ ext{theoretical yield}} imes 100\%

Example 5.14

  • Task: Determine the percent yield of glucose from the reaction where 12.0 g CO2 reacts with 8.00 g H2O to yield 7.50 g C6H12O6.

  • Chemical Equation:
    6 CO2(g) + 6 H2O(l) → C6H{12}O6(aq) + 6 O2(g)

Section 5.4 Review

  • Key Points:

    • Theoretical yield, actual yield, and percent yield are crucial components of understanding chemical reactions.

    • Factors reducing actual yield include competing reactions, residual product, and incomplete reactions.

    • Percent yield calculation method illustrated with Equation 5.1.

Section 5.5: Definition and Uses of Molarity

  • Objective: Define molarity and utilize it to solve concentration problems.

Concentration Concepts

  • Definitions:

    • A solution is a homogeneous mixture of two or more substances. The major component is the solvent and the substance dissolved is the solute.

    • Concentration expresses how much of a solute is present in a given volume of solvent or solution.

    • Solutions can be concentrated (high solute) or dilute (low solute).

Molarity Definition

  • Molarity (M): Defined as the number of moles of solute per liter of solution: ext{molarity} = rac{ ext{moles of solute}}{ ext{liters of solution}}

    • Example: 0.50 M NaCl denotes 0.50 moles of NaCl in each 1.0 L of solution.

Example 5.16

  • Task: Calculate molarity of a solution with 7.50 mol of formaldehyde in 1.50 L of solution.

Molarity as a Conversion Factor

  • Molarity can also be defined as the number of millimoles of solute per milliliter of solution:
    M = rac{ ext{millimoles of solute}}{ ext{milliliters of solution}}

Example 5.17

  • Task: Calculate the molarity of a solution with 1.25 mmol solute in 60.0 mL of solution.

Example 5.18

  • Task: Calculate the molarity of a solution with 7.50 g of methanol in 50.0 mL of solution.

Section 5.5 Review

  • Key Takeaways:

    • Molarity is a crucial concept for understanding solution concentrations and converting between moles and volume.

    • Molarity can be expressed in various ways, assisting in calculations.

Section 5.6: Molarities of Ions

  • Objective: Extend molarity definition to individual ions in ionic solution.

Ionic Solutions

  • Strong electrolytes dissociate into ions in water.

  • Molarity of any ion equals the number of moles of that ion per liter of solution.

  • Example for AlCl3: In a 1.0 M solution, molarity for constituents is 1.0 M Al³⁺ and 3.0 M Cl⁻.

Example 5.25

  • Task: Calculate concentration of ions in a mixed solution made by diluting NaCl and CaCl2.

Section 5.6 Review

  • Key Points:

    • Molarities of ions can be calculated based on the molarity of the electrolyte and its formula ratios.

    • The subscripts in chemical formulas directly relate to the molar ratios of ions present in the solution.

Section 5.7: Calculations Involving Other Quantities

  • Objective: Interpret various terms and convert quantities in chemical reactions.

Conversion Methods

  • Different ways to convert to moles include:

    • Mass to moles via molar mass.

    • Volume to moles via molarity.

    • Volume to mass to moles via density and molar mass.

    • Moles from number of particles via Avogadro’s number.

Example 5.27

  • Task: Determine the number of SO2 molecules needed to produce 0.751 mol SO3.

  • Chemical Equation:
    2 SO2(g) + O2(g) → 2 SO_3(g)

Section 5.8: Calculations with Net Ionic Equations

  • Objective: Work with net ionic equations to understand stoichiometry in reactions.

Net Ionic Equations

  • Net ionic equations represent mole ratios of reactants/products.

  • Spectator ions must be considered for mass calculations.

    • Example: Reaction of AgNO3 with NaCl provides context to use actual ions in calculations.

Example 5.30

  • Task a: Determine mass of Ba(NO3)2 required to provide 0.750 mol Ba²⁺.

  • Task b: Determine mass of Ba(ClO3)2 required for the same cation quantity.

Section 5.9: Titration

  • Objective: Use titration to determine concentration of unknown solutions.

Titration Process

  • Titration is a laboratory method for determining moles of a substance in an aqueous solution by mixing with a standard solution of known concentration.

    • Steps:

    1. The titrant (standard solution) is added to the sample solution using a buret.

    2. An indicator signals the endpoint when the reaction approaches equivalence.

    3. Volume of titrant added is measured.

Example 5.31

  • Task: Calculate concentration of HCl titrated with NaOH. Initial and final volume readings provided.

Section 5.9 Review

  • Key Points:

    • Titration is fundamental in quantifying concentrations, especially in acid-base reactions.

    • The number of moles used is calculated from the reagent’s volume and known concentration; this information is useful for deducing unknown concentrations or masses.

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