Stoichiometry Chapter 11 Flashcards

Stoichiometry Overview

• The study of quantitative relationships in chemical reactions, involving reactants and products.

Mass Relationships

• Mass of reactants equals mass of products in a reaction.

Conservation of Atoms

• The number of atoms for each element must be equal on both sides of the reaction equation.

Molar Relationships

• The number of moles for each element must also be equal on both sides of the equation.

Mole Ratio

• Derived from the coefficients of balanced chemical equations, used for conversions.

Example of Mole Ratios

• For the equation 4Al + 3O₂ → 2Al₂O₃, various mole ratios can be established:
  • 4 moles Al : 3 moles O₂
  • 3 moles O₂ : 2 moles Al₂O₃

Mole to Mole Conversion

• Determine moles of an unknown substance using the formula: Moles of Unknown = Moles of Known x Mole Ratio.

Example Problem: O₂ to Al₂O₃ Conversion

• From the equation, if 3.75 moles O₂ are given, 2.50 moles Al₂O₃ are produced.

Example Problem: C₃H₈ to CO₂ Conversion

• Burning 7 moles of C₃H₈ produces 21 moles of CO₂.

Mass to Mass Conversion

• To convert from mass of the Known to the mass of the Unknown, follow: Mass of Known / Molar Mass of Known x Mole Ratio x Molar Mass of Unknown.

Example Problem: NH₄NO₃ to H₂O

• From 50 g of NH₄NO₃, 22.5 g of H₂O are produced.

Mole to Mass Conversion

• Calculated using: Moles Known x Mole Ratio x Molar Mass of Unknown.

Mass to Mole Conversion

• Achieved through: Mass Known / Molar Mass Known x Mole Ratio.

Limiting and Excess Reactants

• Limiting reactant is the one that produces the least amount of product.
• Excess reactant remains un-reacted after the reaction.

Limiting Reactant Problem Example

• In the reaction S₈ + 4Cl₂ → 4S₂Cl₂ with 200 g S₈ and 100 g Cl₂:
  • Perform separate calculations to determine S₂Cl₂ production from each reactant.

Mass-to-Mass Calculation Example

• For sulfur, 421 g of S₂Cl₂ can be produced.
• For chlorine, only 190.4 g of S₂Cl₂ can be produced, making Cl₂ the limiting reactant.

Excess Reactant Calculation

• Remaining S₈ after reaction:
  • 200 g S₈ supplied, 93.6 g used, leaving 106.4 g in excess.

Percent Yield

• Ratio of actual amount produced to the expected (theoretical) amount, expressed as a percentage.

Theoretical Yield

• Expected production based on stoichiometric calculations.

Actual Yield

• The amount of product produced during an experiment.

Percent Yield Equation

• Percent Yield = (Actual Yield ÷ Theoretical Yield) × 100.

Percent Yield Example Problem: Ag₂CrO₄

• Calculating yield from reaction K₂CrO₄ + 2AgNO₃ → Ag₂CrO₄ involves finding theoretical yield from limiting reactant AgNO₃ first.
• Given 0.500 g of AgNO₃, the theoretical yield calculated is 0.488 g.
• Actual yield of Ag₂CrO₄ is 0.455 g, leading to a percent yield of 93.2%.