Limiting Reactant Practice Problems And Answers

Limiting Reactant Practice Problems and Answers: Mastering Stoichiometry

Stoichiometry, the heart of quantitative chemistry, often presents the challenge of determining the limiting reactant. This crucial concept dictates the maximum amount of product that can be formed in a chemical reaction. Understanding how to identify the limiting reactant is essential for predicting reaction yields and optimizing chemical processes. This comprehensive guide will provide a range of practice problems, complete with detailed solutions, to solidify your understanding of this fundamental concept. We'll explore different approaches to solving these problems, building your confidence and problem-solving skills.

What is a Limiting Reactant?

Before diving into the problems, let's refresh our understanding of limiting reactants. In a chemical reaction involving multiple reactants, the limiting reactant is the reactant that is completely consumed first. Once this reactant is used up, the reaction stops, regardless of how much of the other reactants remains. The other reactants are considered to be in excess. The limiting reactant directly determines the theoretical yield of the product – the maximum amount of product that can be formed based on the stoichiometry of the reaction.

Problem Solving Strategies

Solving limiting reactant problems involves a series of steps:

1. Balanced Chemical Equation: Always start with a correctly balanced chemical equation. This equation provides the crucial mole ratios between reactants and products.

2. Convert to Moles: Convert the given masses (or volumes if dealing with gases at STP) of each reactant into moles using their respective molar masses.

3. Mole Ratio Comparison: Use the mole ratios from the balanced equation to determine how many moles of each reactant are needed to completely react with the available moles of the other reactant(s).

4. Identify the Limiting Reactant: The reactant that produces the least amount of product is the limiting reactant. This reactant will be completely consumed before the others.

5. Calculate Theoretical Yield: Once the limiting reactant is identified, use its moles and the mole ratio from the balanced equation to calculate the theoretical yield of the desired product.

Practice Problems

Let's put these strategies into action with a series of practice problems of increasing complexity.

Problem 1: Simple Stoichiometry

Question: Consider the reaction: 2H₂ + O₂ → 2H₂O. If 2 moles of hydrogen gas (H₂) react with 1 mole of oxygen gas (O₂), which reactant is limiting, and how many moles of water (H₂O) are produced?

Solution:

1. Balanced Equation: The equation is already balanced.

2. Moles: We have 2 moles of H₂ and 1 mole of O₂.

3. Mole Ratio: According to the balanced equation, 2 moles of H₂ react with 1 mole of O₂. We have enough O₂ to react with all the H₂.

4. Limiting Reactant: Let's check:

   • 2 moles H₂ × (1 mole O₂ / 2 moles H₂) = 1 mole O₂ needed
   • We have 1 mole O₂, which is exactly what's needed. Neither reactant is limiting in this specific case; they are present in the stoichiometric ratio.

5. Theoretical Yield:

   • 2 moles H₂ × (2 moles H₂O / 2 moles H₂) = 2 moles H₂O

Answer: Neither reactant is limiting. 2 moles of water are produced.

Problem 2: Identifying the Limiting Reactant

Question: Given the reaction: N₂ + 3H₂ → 2NH₃. If 14 grams of nitrogen gas (N₂) react with 6 grams of hydrogen gas (H₂), which is the limiting reactant?

Solution:

1. Balanced Equation: The equation is already balanced.

2. Moles:

   • Moles of N₂: 14 g N₂ × (1 mol N₂ / 28 g N₂) = 0.5 mol N₂
   • Moles of H₂: 6 g H₂ × (1 mol H₂ / 2 g H₂) = 3 mol H₂

3. Mole Ratio:

   • From the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂.
   • 0.5 mol N₂ × (3 mol H₂ / 1 mol N₂) = 1.5 mol H₂ needed

4. Limiting Reactant: We have 3 moles of H₂, but only 1.5 moles are needed. Therefore, H₂ is in excess, and N₂ is the limiting reactant.

Problem 3: Calculating Theoretical Yield

Question: Continuing from Problem 2, what is the theoretical yield of ammonia (NH₃) in grams?

Solution:

1. Limiting Reactant: We established that N₂ is the limiting reactant (0.5 moles).

2. Mole Ratio: From the balanced equation, 1 mole of N₂ produces 2 moles of NH₃.

3. Moles of NH₃: 0.5 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1 mol NH₃

4. Mass of NH₃: 1 mol NH₃ × (17 g NH₃ / 1 mol NH₃) = 17 g NH₃

Answer: The theoretical yield of ammonia is 17 grams.

Problem 4: Involving More Reactants

Question: Consider the reaction: 2Fe + 3Cl₂ → 2FeCl₃. If 11.2 grams of iron (Fe) react with 21.3 grams of chlorine gas (Cl₂), what is the limiting reactant, and what mass of iron(III) chloride (FeCl₃) is produced?

Solution:

1. Balanced Equation: The equation is balanced.

2. Moles:

   • Moles of Fe: 11.2 g Fe × (1 mol Fe / 55.85 g Fe) ≈ 0.200 mol Fe
   • Moles of Cl₂: 21.3 g Cl₂ × (1 mol Cl₂ / 70.9 g Cl₂) ≈ 0.300 mol Cl₂

3. Mole Ratio:

   • 0.200 mol Fe × (3 mol Cl₂ / 2 mol Fe) = 0.300 mol Cl₂ needed

4. Limiting Reactant: We need 0.300 mol Cl₂, and we have exactly 0.300 mol Cl₂. Therefore, neither reactant is in excess. Both are in the stoichiometric ratio.

5. Theoretical Yield:

   • 0.200 mol Fe × (2 mol FeCl₃ / 2 mol Fe) = 0.200 mol FeCl₃
   • 0.200 mol FeCl₃ × (162.2 g FeCl₃ / 1 mol FeCl₃) ≈ 32.4 g FeCl₃

Answer: Neither reactant is limiting. Approximately 32.4 grams of FeCl₃ are produced.

Problem 5: Dealing with Excess Reactant

Question: For the reaction: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. If 20 grams of sodium hydroxide (NaOH) react with 30 grams of sulfuric acid (H₂SO₄), what is the limiting reactant, and how many grams of sodium sulfate (Na₂SO₄) are produced?

Solution:

1. Balanced Equation: The equation is balanced.

2. Moles:

   • Moles of NaOH: 20 g NaOH × (1 mol NaOH / 40 g NaOH) = 0.5 mol NaOH
   • Moles of H₂SO₄: 30 g H₂SO₄ × (1 mol H₂SO₄ / 98 g H₂SO₄) ≈ 0.306 mol H₂SO₄

3. Mole Ratio:

   • 0.5 mol NaOH × (1 mol H₂SO₄ / 2 mol NaOH) = 0.25 mol H₂SO₄ needed

4. Limiting Reactant: We need 0.25 mol H₂SO₄, but we have 0.306 mol H₂SO₄. Therefore, H₂SO₄ is in excess, and NaOH is the limiting reactant.

5. Theoretical Yield:

   • 0.5 mol NaOH × (1 mol Na₂SO₄ / 2 mol NaOH) = 0.25 mol Na₂SO₄
   • 0.25 mol Na₂SO₄ × (142 g Na₂SO₄ / 1 mol Na₂SO₄) = 35.5 g Na₂SO₄

Answer: NaOH is the limiting reactant. Approximately 35.5 grams of Na₂SO₄ are produced.

Advanced Problem Solving Techniques

For more complex scenarios, consider these additional techniques:

• Percent Yield: Often, the actual yield of a reaction is less than the theoretical yield. The percent yield accounts for this difference: (Actual Yield / Theoretical Yield) × 100%.

• Limiting Reactant with Multiple Products: If the reaction produces multiple products, you'll need to calculate the theoretical yield for each product based on the limiting reactant.

• Sequential Reactions: Some reactions involve multiple steps. Identify the limiting reactant for each step to determine the overall limiting reactant and final product yield.

Conclusion

Mastering limiting reactant calculations is crucial for success in chemistry. By systematically applying the steps outlined above and practicing with various problems, you can build your confidence and proficiency in stoichiometric calculations. Remember to always start with a balanced chemical equation and pay close attention to mole ratios. The problems presented here provide a solid foundation; continue practicing to develop a deeper understanding of this fundamental concept. Remember that consistent practice and a thorough understanding of the underlying principles are key to success in solving any stoichiometry problem.