Stoichiometry_StudyGuide

Chapter 9: Chemical Quantities

9.1 Information Given by Chemical Equations

• Understanding Molecular and Mass Information in a Balanced Equation

  • A balanced chemical equation provides the amounts of reactant and product molecules involved in a reaction.

  • Coefficients in the equation indicate relative numbers of molecules.

• Equation Example: C2H5OH + 3O2 → 2CO2 + 3H2O

  • Balanced equation: All reactant atoms are accounted for in the products.

  • 1 molecule of ethanol reacts with 3 molecules of oxygen to produce:

    • 2 molecules of carbon dioxide (CO2)

    • 3 molecules of water (H2O)

  • This also corresponds to moles: 1 mol of ethanol reacts with 3 mol of O2 to yield 2 mol CO2 and 3 mol H2O.

9.2 Mole–Mole Relationships

• Relationship of Reactants to Products

  • Balanced equations allow prediction of the moles of products from given moles of reactants.

  • For example:

    • 2 moles of H2O yield 2 moles of H2 and 1 mole of O2.

    • 4 mol H2O yields 4 mol H2 and 2 mol O2.

• Mole Ratio

  • Enables conversions between moles of one substance and moles of another in the balanced equation.

• Example Problem:

• Equation: Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s)

  • Determine the moles of NaF produced from 3.50 moles of Na reacting with excess Na2SiF6.

  • Know that the balanced equation informs this calculation.

• Exercise 1: Predict CO2 moles from burning 3.74 moles of propane (C3H8) in excess O2.

  • Equation: C3H8 + 5O2 → 3CO2 + 4H2O

  • Options: a) 11.2 b) 7.48 c) 3.74 d) 1.25.

9.3 Mass Calculations

• Steps for Calculating Masses

  • Convert mass of reactants/products to moles.

  • Use balanced equations for mole ratios.

  • Compute moles of desired reactant/product.

  • Convert moles back to mass as needed.

• Stoichiometry

  • The process of using a balanced chemical equation to calculate masses of reactants and products.

• Example Problem:

  • For the reaction: Cr(s) + O2(g) → Cr2O3(s)

    • What grams of Cr2O3 are produced from 15.0 g of Cr and excess O2?

    • Molar Masses: Cr = 52.00 g/mol and O = 16.00 g/mol.

• Example Calculation Steps:

  • Convert Cr mass to moles.

  • Use mole ratio to find moles of Cr2O3.

  • Convert moles of Cr2O3 to grams.

• Exercise 2:

  • Sn(s) + 2HF(aq) → SnF2(aq) + H2(g)

  • How many grams of SnF2 from 55.0 g HF?

  • Options: a) 431 g b) 215 g c) 72.6 g d) 1.37 g.

• Exercise 3:

  • PCl3 + 3H2O → H3PO3 + 3HCl

  • Mass of H2O needed for 20.0 g of PCl3?

  • Options: a) 7.87 b) 0.875 c) 5.24 d) 2.62.

9.4 The Concept of Limiting Reactants

• Stoichiometric Mixture

  • Contains reactant amounts that correspond to a balanced equation.

  • Example: N2(g) + 3H2(g) → 2NH3(g)

• Limiting Reactant Definition

  • The one that runs out first, limiting product formation.

  • Example: H2 is the limiting reactant.

9.5 Calculations Involving a Limiting Reactant

• Identifying Limiting Reactants

  • Determine the limiting reactant to calculate product amounts.

  • Reaction: CH4 + H2O → 3H2 + CO

  • H2O is consumed first, making it the limiting reactant; CH4 is in excess.

• Steps for Solving Stoichiometry Problems

  • Convert known reactant masses to moles.

  • Compute moles of product possible from each reactant based on ratios.

  • Select the least amount produced as the limiting factor.

  • Convert to grams if needed based on molar mass.

• Concept Checks

  • Which reaction mixture produces most product in: 2H2 + O2 → 2H2O? (multiple options)

  • If 3.0 moles of A react with 6.0 moles of B in: A + 3B → 4C, determine excess reactant (answers depend on moles consumed).

9.6 Percent Yield

• Percent Yield Explanation

  • Indicator of reaction efficiency comparing actual yield to theoretical yield.

  • Theoretical Yield: Maximum expected amount from complete reactant consumption.

  • Actual Yield: Practical amount obtained, often less than theoretical due to various factors.

• Calculating Percent Yield

  • Percent yield = (Actual Yield / Theoretical Yield) × 100%

• Example Problem:

  • Reaction of A and B yields mass C.

  • Determine percent yield given theoretical (8.33g) vs. actual (7.23g).

• Exercise 6:

  • 2S + 3O2 → 2SO3; find percent yield of 1.50 g of SO3 produced from 1.00 g O2 (options available).

• Exercise 7:

  • Reaction P4 + 6F2 → 4PF3; determine mass of P4 for 85.0 g of PF3 with a 64.9% yield (multiple choice available).

Chapter 9 discusses Chemical Quantities, highlighting how balanced chemical equations convey information about molecular and mass relationships in chemical reactions.

• 9.1 Information Given by Chemical Equations: A balanced equation indicates the amounts of reactants and products, with coefficients signifying the relative number of molecules (e.g., C2H5OH + 3O2 → 2CO2 + 3H2O).

• 9.2 Mole–Mole Relationships: Allows for predicting moles of products based on given moles of reactants using mole ratios (e.g., 2 moles of H2O yield 2 moles of H2).

• 9.3 Mass Calculations: Discusses converting mass to moles and vice versa, using stoichiometry to calculate masses of reactants/products based on balanced equations.

• 9.4 The Concept of Limiting Reactants: Defines limiting reactants as those that run out first, affecting product yield.

• 9.5 Calculations Involving a Limiting Reactant: Details steps for identifying limiting reactants and calculating product amounts.

• 9.6 Percent Yield: Explains the concept of percent yield, comparing actual yield to theoretical yield to assess reaction efficiency.