Stoichiometry and Limiting Reagents

Stoichiometry Applications

Mole Ratios

• Balanced reactions provide mole ratios of reactants consumed to products generated.
• Mole ratios can be generated for:
  • Reactant to product
  • Reactant to reactant
  • Product to product
• These ratios are based on stoichiometric coefficients.
• Example: Formation of water: $2H<em>2 + O</em>2 \rightarrow 2H_2O$
  • For every 2 moles of $H<em>2$ consumed, 2 moles of $H</em>2O$ are produced.
  • For every 1 mole of $O<em>2$ consumed, 2 moles of $H</em>2O$ are produced.
  • $H<em>2$ is consumed twice as fast as $O</em>2$.
• Stoichiometry problems involve unit conversions.

Example Problem: Calcium Chloride and Silver Chloride

• Problem: How many grams of $CaCl_2$ are needed to prepare 71.7 g of $AgCl$?
• Balanced Equation: $CaCl<em>2(aq) + 2AgNO</em>3(aq) \rightarrow Ca(NO<em>3)</em>2(aq) + 2AgCl(s)$
• From the balanced equation:
  • 1 mole of $CaCl_2$ reacts to yield 2 moles of $AgCl$.
• Molar masses:
  • $CaCl_2$: 111.1 g/mol
  • $AgCl$: 143.4 g/mol
• Calculation:
  • $71.7 g \ AgCl * (\frac{1 \ mol \ AgCl}{143.4 \ g \ AgCl}) * (\frac{1 \ mol \ CaCl<em>2}{2 \ mol \ AgCl}) * (\frac{111.1 \ g \ CaCl</em>2}{1 \ mol \ CaCl<em>2}) = 27.775 \ g \ CaCl</em>2$
  • $\approx 27.8 \ g \ CaCl_2$

Limiting Reagent

• Reactants are rarely added in exact stoichiometric proportions.
• Limiting Reagent: The reactant that is completely consumed, thus limiting the amount of product formed.
• Excess Reagents: Reactants remaining after the limiting reagent is used up.
• Principles for determining the limiting reagent:
  1. Comparisons of reactants must be in moles.
  2. The stoichiometric ratios and absolute mole quantities determine the limiting reagent, not just absolute mole quantities.

Example Problem: Iron and Sulfur

• Problem: If 27.9 g of Fe react with 24.1 g of S to produce FeS, what is the limiting reagent? How many grams of excess reagent remain?
• Balanced Equation: $Fe + S \rightarrow FeS$
• Calculations:
  • Moles of Fe: $27.9 \ g \ Fe * (\frac{1 \ mol \ Fe}{55.8 \ g \ Fe}) = 0.5 \ mol \ Fe$
  • Moles of S: $24.1 \ g \ S * (\frac{1 \ mol \ S}{32.1 \ g \ S}) = 0.75 \ mol \ S$
• Since the reaction requires a 1:1 mole ratio of Fe to S, Fe is the limiting reagent because there are only 0.5 moles of Fe.
• 0.5 moles of Fe will react with 0.5 moles of S, leaving 0.25 moles of S in excess.
• Mass of excess S: $0.25 \ mol \ S * (\frac{32.1 \ g \ S}{1 \ mol \ S}) = 8.025 \ g \ S \approx 8 \ g \ S$

Yield

• Theoretical Yield: The maximum amount of product predicted from the balanced equation, assuming complete consumption of the limiting reactant, no side reactions, and complete product collection.
• Actual Yield: The amount of product actually obtained from the reaction.
• Percent Yield: The ratio of actual yield to theoretical yield, multiplied by 100%.
• Formula: $Percent \ Yield = (\frac{Actual \ Yield}{Theoretical \ Yield}) * 100\%$

Example Problem: Zinc and Copper

• Problem: What is the percent yield for a reaction in which 28 g of Cu is produced by reacting 32.7 g of Zn in excess $CuSO_4$ solution?
• Balanced Equation: $Zn(s) + CuSO<em>4(aq) \rightarrow Cu(s) + ZnSO</em>4(aq)$
• Calculate the theoretical yield of Cu:
  • $32.7 \ g \ Zn * (\frac{1 \ mol \ Zn}{65.4 \ g \ Zn}) * (\frac{1 \ mol \ Cu}{1 \ mol \ Zn}) * (\frac{63.5 \ g \ Cu}{1 \ mol \ Cu}) = 31.8 \ g \ Cu$
  • Therefore, the theoretical yield is 31.8 g Cu.
• Calculate percent yield:
  • $(\frac{28 \ g \ Cu}{31.8 \ g \ Cu}) * 100\% = 87.5\%$