Limiting Reactant Practice Problems With Answers

Limiting Reactant Practice Problems with Answers: Mastering Stoichiometry

Stoichiometry, the section of chemistry dealing with the quantitative relationships between reactants and products in chemical reactions, often presents the challenge of determining the limiting reactant. This crucial concept dictates the maximum amount of product that can be formed in a reaction. Understanding how to identify the limiting reactant is essential for success in chemistry, and mastering this skill requires consistent practice. This comprehensive guide provides a range of practice problems of varying difficulty, complete with detailed solutions, to solidify your understanding of this critical topic.

What is a Limiting Reactant?

Before diving into the problems, let's revisit the fundamental concept. In a chemical reaction, reactants are the starting materials, and products are the substances formed. The limiting reactant (or limiting reagent) is the reactant that is completely consumed first in a chemical reaction, thus limiting the amount of product that can be formed. Once the limiting reactant is used up, the reaction stops, regardless of how much of the other reactants remains. The reactants that are left over are called excess reactants.

Understanding Mole Ratios

The key to solving limiting reactant problems lies in understanding mole ratios. These ratios are derived directly from the balanced chemical equation. The coefficients in the balanced equation represent the relative number of moles of each reactant and product involved in the reaction. For example, in the balanced equation:

2H₂ + O₂ → 2H₂O

The mole ratio of hydrogen to oxygen is 2:1, meaning that for every 2 moles of hydrogen that react, 1 mole of oxygen is required. Similarly, the mole ratio of hydrogen to water is 2:2 (or 1:1).

Steps to Solve Limiting Reactant Problems

To determine the limiting reactant, follow these steps:

1. Write and balance the chemical equation: Ensure the equation accurately represents the reaction and is balanced to maintain the law of conservation of mass.

2. Convert given quantities to moles: Use molar mass to convert the given masses of reactants (usually in grams) into moles.

3. Determine the mole ratio: Use the balanced equation's coefficients to establish the mole ratio between the reactants.

4. Compare mole ratios to determine the limiting reactant: Compare the actual mole ratio of reactants to the stoichiometric mole ratio from the balanced equation. The reactant that has fewer moles relative to its stoichiometric coefficient is the limiting reactant.

5. Calculate the theoretical yield: Use the moles of the limiting reactant and the mole ratio from the balanced equation to calculate the moles of product formed. Then, convert moles of product to grams using its molar mass.

Practice Problems

Now let's put these steps into action with several practice problems. Remember to show your work!

Problem 1:

Consider the reaction: N₂(g) + 3H₂(g) → 2NH₃(g)

If 28 grams of nitrogen gas (N₂) react with 12 grams of hydrogen gas (H₂), which is the limiting reactant, and what is the theoretical yield of ammonia (NH₃) in grams?

Problem 2:

For the reaction: 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s)

10.0 grams of iron (Fe) react with 15.0 grams of chlorine gas (Cl₂). Determine the limiting reactant and calculate the theoretical yield of iron(III) chloride (FeCl₃) in grams.

Problem 3:

The reaction between sodium carbonate (Na₂CO₃) and hydrochloric acid (HCl) produces sodium chloride (NaCl), water (H₂O), and carbon dioxide (CO₂):

Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

If 21.2 grams of sodium carbonate react with 30.0 grams of hydrochloric acid, what is the limiting reactant, and what is the theoretical yield of carbon dioxide in grams?

Problem 4 (Slightly More Advanced):

The combustion of propane (C₃H₈) is represented by the following balanced equation:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

If 10.0 grams of propane react with 30.0 grams of oxygen gas, what is the limiting reactant? Calculate the mass of water produced. Additionally, calculate the mass of the excess reactant remaining after the reaction is complete.

Problem 5 (Advanced - Involving Percent Yield):

Consider the reaction: 2Mg(s) + O₂(g) → 2MgO(s)

If 4.86 grams of magnesium (Mg) react with excess oxygen to produce 7.50 grams of magnesium oxide (MgO), what is the percent yield of the reaction? (Hint: First find the theoretical yield).

Answers and Detailed Solutions

Problem 1:

1. Balanced Equation: The equation is already balanced.

2. Moles:

   • Moles of N₂ = (28 g) / (28 g/mol) = 1.0 mol
   • Moles of H₂ = (12 g) / (2 g/mol) = 6.0 mol

3. Mole Ratio: From the balanced equation, the mole ratio of N₂ to H₂ is 1:3. For every 1 mole of N₂, 3 moles of H₂ are needed.

4. Limiting Reactant: We have 1.0 mol of N₂. To react completely, this would require 3 * 1.0 mol = 3.0 mol of H₂. Since we have 6.0 mol of H₂, H₂ is in excess, and N₂ is the limiting reactant.

5. Theoretical Yield: 1.0 mol of N₂ produces 2.0 mol of NH₃ (from the mole ratio). The mass of NH₃ produced is (2.0 mol) * (17 g/mol) = 34 g.

Problem 2:

1. Balanced Equation: The equation is already balanced.

2. Moles:

   • Moles of Fe = (10.0 g) / (55.85 g/mol) = 0.179 mol
   • Moles of Cl₂ = (15.0 g) / (70.9 g/mol) = 0.212 mol

3. Mole Ratio: The mole ratio of Fe to Cl₂ is 2:3. For every 2 moles of Fe, 3 moles of Cl₂ are needed.

4. Limiting Reactant: 0.179 mol of Fe requires (3/2) * 0.179 mol = 0.269 mol of Cl₂. Since we only have 0.212 mol of Cl₂, Cl₂ is the limiting reactant.

5. Theoretical Yield: 0.212 mol of Cl₂ produces (2/3) * 0.212 mol = 0.141 mol of FeCl₃. The mass of FeCl₃ is (0.141 mol) * (162.2 g/mol) = 22.9 g.

Problem 3:

1. Balanced Equation: The equation is already balanced.

2. Moles:

   • Moles of Na₂CO₃ = (21.2 g) / (105.99 g/mol) = 0.200 mol
   • Moles of HCl = (30.0 g) / (36.46 g/mol) = 0.823 mol

3. Mole Ratio: The mole ratio of Na₂CO₃ to HCl is 1:2.

4. Limiting Reactant: 0.200 mol of Na₂CO₃ requires 2 * 0.200 mol = 0.400 mol of HCl. Since we have 0.823 mol of HCl, HCl is in excess, and Na₂CO₃ is the limiting reactant.

5. Theoretical Yield: 0.200 mol of Na₂CO₃ produces 0.200 mol of CO₂. The mass of CO₂ is (0.200 mol) * (44.01 g/mol) = 8.80 g.

Problem 4:

1. Balanced Equation: The equation is already balanced.

2. Moles:

   • Moles of C₃H₈ = (10.0 g) / (44.1 g/mol) = 0.227 mol
   • Moles of O₂ = (30.0 g) / (32.0 g/mol) = 0.938 mol

3. Mole Ratio: The mole ratio of C₃H₈ to O₂ is 1:5.

4. Limiting Reactant: 0.227 mol of C₃H₈ requires 5 * 0.227 mol = 1.135 mol of O₂. Since we only have 0.938 mol of O₂, O₂ is the limiting reactant.

5. Mass of Water Produced: 0.938 mol of O₂ produces (4/5) * 0.938 mol = 0.750 mol of H₂O. The mass of H₂O is (0.750 mol) * (18.02 g/mol) = 13.5 g.

6. Mass of Excess Reactant: The moles of C₃H₈ reacted are (1/5) * 0.938 mol = 0.188 mol. The moles of C₃H₈ remaining are 0.227 mol - 0.188 mol = 0.039 mol. The mass of C₃H₈ remaining is (0.039 mol) * (44.1 g/mol) = 1.72 g.

Problem 5:

1. Balanced Equation: The equation is already balanced.

2. Moles of Mg: Moles of Mg = (4.86 g) / (24.31 g/mol) = 0.200 mol

3. Theoretical Yield: 0.200 mol of Mg produces 0.200 mol of MgO. The theoretical yield of MgO is (0.200 mol) * (40.31 g/mol) = 8.06 g.

4. Percent Yield: Percent yield = (actual yield / theoretical yield) * 100% = (7.50 g / 8.06 g) * 100% = 93.1%.

These problems offer a comprehensive range of difficulty levels, helping you solidify your understanding of limiting reactants. Remember, consistent practice is key to mastering this crucial concept in stoichiometry. By working through these problems and understanding the underlying principles, you'll build a strong foundation for success in your chemistry studies. Remember to always double-check your calculations and ensure your units are consistent throughout the problem-solving process.