Stoichiometry: Limiting Reagents and Percent Yield

Limiting Reagents

• Concept Definition: Limiting reagents are substances in a chemical reaction that determine the maximum amount of product that can be formed because they are consumed first.
• Analogy (Cake Recipe):
  • Recipe: Each cake requires $2$ eggs, $2$ cups of flour, $1$ cup of sugar.
  • Available: $10$ eggs, $6$ cups of flour, $4$ cups of sugar.
  • Calculation: You can make:
    • $10 ext{ eggs} / 2 ext{ eggs/cake} = 5 ext{ cakes}$ (if eggs were the only limit)
    • $6 ext{ cups flour} / 2 ext{ cups flour/cake} = 3 ext{ cakes}$ (if flour were the only limit)
    • $4 ext{ cups sugar} / 1 ext{ cup sugar/cake} = 4 ext{ cakes}$ (if sugar were the only limit)
  • Conclusion: The maximum number of cakes you can make is $3$, because flour is the limiting reagent, running out first. Eggs and sugar will be left over.
• Application to Chemical Reactions:
  • The same concept applies to chemical reactions, but requires unit conversions rather than simple counting.
  • Method: Perform two separate stoichiometry calculations, one for each reactant, to determine the maximum amount of product that can be formed from each.
  • The reactant that yields the least amount of product is the limiting reagent, and its corresponding product amount is the maximum possible yield.
• Example 1: Reaction: N2 (g) + 3H2 (g) ightarrow 2NH_3 (g)
  • Molar Masses: $N<em>2$ = $28 ext{ g/mol}$, $H</em>2$ = $2 ext{ g/mol}$, $NH_3$ = $17 ext{ g/mol}$
  • Given: $10 ext{ g}$ of $N<em>2$ and $4.5 ext{ g}$ of $H</em>2$
  • Problem: How many grams of $NH_3$ can be made?
  • Calculation with $N<em>2$:
    $10 ext{ g } N</em>2 imes \frac{1 ext{ mol } N<em>2}{28 ext{ g } N</em>2} imes \frac{2 ext{ mol } NH<em>3}{1 ext{ mol } N</em>2} imes \frac{17 ext{ g } NH<em>3}{1 ext{ mol } NH</em>3} = 12.14 ext{ g } NH_3$
  • Calculation with $H<em>2$:
    $4.5 ext{ g } H</em>2 imes \frac{1 ext{ mol } H<em>2}{2 ext{ g } H</em>2} imes \frac{2 ext{ mol } NH<em>3}{3 ext{ mol } H</em>2} imes \frac{17 ext{ g } NH<em>3}{1 ext{ mol } NH</em>3} = 25.5 ext{ g } NH_3$
  • Conclusion: The smaller product amount is $12.14 ext{ g } NH<em>3$, meaning $N</em>2$ is the limiting reagent (like flour in the cake analogy). This is the maximum amount of $NH_3$ that can be produced.
• Example 2: Reaction: FeCl3 (aq) + 3NaOH (aq) ightarrow Fe(OH)3 (s) + 3NaCl (aq)
  • Molar Masses: $FeCl<em>3$ = $162.2 ext{ g/mol}$, $NaOH$ = $40 ext{ g/mol}$, $Fe(OH)</em>3$ = $106.8 ext{ g/mol}$, $NaCl$ = $58.45 ext{ g/mol}$
  • Given: $2 ext{ g}$ of $FeCl_3$ and $6 ext{ g}$ of $NaOH$
  • Problem: How many grams of $Fe(OH)_3$ can be made (or identify the limiting reagent)?
  • Method: Set up two stoichiometry problems, one starting with $2 ext{ g } FeCl<em>3$ to $Fe(OH)</em>3$ and another starting with $6 ext{ g } NaOH$ to $Fe(OH)_3$. The reactant that yields the least product is the limiting reagent.
  • Note: It is not possible to determine the limiting reagent by simply looking at masses or coefficients; full calculations are required.

Percent Yield

• Theoretical Yield:
  • Definition: The maximum amount of product that can be produced from a given amount of reactant, assuming the reaction proceeds perfectly (ideal conditions, $100$ efficiency, all molecules react as expected).
  • Calculation: All stoichiometry problems previously worked (grams to moles, moles to moles, moles to grams) calculate theoretical yields.
• Actual Yield:
  • Definition: The amount of product actually obtained and measured in a laboratory experiment.
  • Discrepancy with Theoretical Yield: Actual yield rarely matches theoretical yield due to:
    • Product Loss: Mechanical losses during transfers, stirring, filtration, or other manipulations.
    • Incomplete Reaction: Not all reactants may convert to products.
    • Side Reactions: Other reactions may occur, consuming reactants or forming unwanted products.
    • Impurities: If the product is not completely dried (e.g., contains residual water), the measured mass might exceed the theoretical yield, potentially leading to a percent yield greater than $100$. This does not mean new matter was created; it indicates impurities were present.
• Percent Yield Formula: $ext{Percent Yield} = \frac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100$
  • Units: The units for actual yield and theoretical yield must be consistent (e.g., both in grams, both in moles, both in molecules).
• Example Calculation: Reaction: 2AgNO3 (aq) + Na2SO4 (aq) ightarrow Ag2SO4 (s) + 2NaNO3 (aq)
  • Molar Masses (calculated as needed for $Ag<em>2SO</em>4$):
    • $Ag$ ($107.9 ext{ g/mol}$) + $N$ ($14 ext{ g/mol}$) + $3 ext{ O}$ ($3 imes 16 ext{ g/mol}$) = $107.9 + 14 + 48 = 169.9 ext{ g/mol } AgNO_3$ (rounded from $169.86$)
    • $2 ext{ Ag}$ ($2 imes 107.9 ext{ g/mol}$) + $S$ ($32.1 ext{ g/mol}$) + $4 ext{ O}$ ($4 imes 16 ext{ g/mol}$) = $215.8 + 32.1 + 64 = 311.9 ext{ g/mol } Ag<em>2SO</em>4$
  • Problem 1: Calculate Theoretical Yield of $Ag<em>2SO</em>4$ if $2 ext{ g}$ of $AgNO<em>3$ reacts. $2 ext{ g } AgNO</em>3 imes \frac{1 ext{ mol } AgNO<em>3}{169.9 ext{ g } AgNO</em>3} imes \frac{1 ext{ mol } Ag<em>2SO</em>4}{2 ext{ mol } AgNO<em>3} imes \frac{311.9 ext{ g } Ag</em>2SO<em>4}{1 ext{ mol } Ag</em>2SO<em>4} = 1.84 ext{ g } Ag</em>2SO_4$
    • This $1.84 ext{ g}$ is the theoretical yield.
  • Problem 2: Calculate Percent Yield if an actual yield of $1.59 ext{ g}$ of $Ag<em>2SO</em>4$ is obtained.
    $ext{Percent Yield} = \frac{1.59 ext{ g}}{1.84 ext{ g}} imes 100 = 86.4$
• Interpretation: A $100$ yield is highly suspicious in real-world experiments. The