Limiting Reagent Practice Problems With Answers

In the realm of chemistry, mastering the concept of the limiting reagent is crucial for accurate stoichiometric calculations and predictions. The limiting reagent dictates the maximum amount of product that can be formed in a chemical reaction. This article delves into various practice problems, equipping you with the knowledge and skills to identify the limiting reagent and calculate theoretical yields. Let's embark on this journey to solidify your understanding of this fundamental chemical concept.

Understanding the Limiting Reagent

The limiting reagent in a chemical reaction is the reactant that is completely consumed first, thereby limiting the amount of product that can be formed. In contrast, the reactant(s) present in excess are termed excess reagents. Identifying the limiting reagent is essential for determining the theoretical yield of a reaction, which represents the maximum amount of product that can be obtained if the reaction proceeds to completion with 100% efficiency.

Steps to Solve Limiting Reagent Problems

Before diving into practice problems, let's outline a systematic approach to solving them:

1. Balance the Chemical Equation: Ensure that the chemical equation is balanced to accurately represent the stoichiometry of the reaction.
2. Convert Given Masses to Moles: Convert the given masses of reactants to moles using their respective molar masses.
3. Determine the Mole Ratio: Calculate the mole ratio of the reactants based on the balanced chemical equation.
4. Identify the Limiting Reagent: Compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation. The reactant with the smaller ratio is the limiting reagent.
5. Calculate the Theoretical Yield: Use the moles of the limiting reagent to calculate the theoretical yield of the product(s).
6. Calculate the Percent Yield (If Required): If the actual yield of the product is given, calculate the percent yield using the formula: Percent Yield = (Actual Yield / Theoretical Yield) x 100%.

Practice Problems with Answers

Let's tackle a series of practice problems to illustrate the application of these steps.

Problem 1:

Consider the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to produce ammonia (NH₃):

N₂(g) + 3H₂(g) → 2NH₃(g)

If you have 28.0 g of N₂ and 6.0 g of H₂, which is the limiting reagent, and what is the theoretical yield of NH₃ in grams?

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Masses to Moles:

   • Moles of N₂ = 28.0 g / 28.02 g/mol = 0.999 mol ≈ 1 mol
   • Moles of H₂ = 6.0 g / 2.02 g/mol = 2.97 mol ≈ 3 mol

3. Determine the Mole Ratio:

   • Mole ratio of N₂ to H₂ = 1 mol N₂ / 3 mol H₂ = 1:3

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of N₂ to H₂ is 1:3.
   • We have approximately 1 mol of N₂ and 3 mol of H₂. The ratio matches the stoichiometric ratio.
   • However, a closer look reveals a slight difference: if all the N₂ reacts, we need 3 moles of H₂, which we have. If all the H₂ reacts, we need 1 mole of N₂, which we also have. The more precise calculation is:
     • To consume 1 mole of N₂, we need 3 moles of H₂. We have 3 moles of H₂, so N₂ is the limiting reagent.
     • To consume 3 moles of H₂, we need 1 mole of N₂. We have 1 mole of N₂, so H₂ is not in excess.
   • Therefore, N₂ is the limiting reagent.

5. Calculate the Theoretical Yield:

   • From the balanced equation, 1 mol of N₂ produces 2 mol of NH₃.
   • Therefore, 1 mol of N₂ will produce 2 mol of NH₃.
   • Mass of NH₃ = 2 mol x 17.03 g/mol = 34.06 g
   • The theoretical yield of NH₃ is approximately 34.06 g.

Problem 2:

What mass of precipitate (AgCl) is formed when 200.0 mL of 0.100 M silver nitrate (AgNO₃) is mixed with 150.0 mL of 0.150 M calcium chloride (CaCl₂)?

2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of AgNO₃ = (0.200 L) x (0.100 mol/L) = 0.0200 mol
   • Moles of CaCl₂ = (0.150 L) x (0.150 mol/L) = 0.0225 mol

3. Determine the Mole Ratio:

   • Mole ratio of AgNO₃ to CaCl₂ = 0.0200 mol / 0.0225 mol = 0.889

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of AgNO₃ to CaCl₂ is 2:1.
   • To consume 0.0225 mol of CaCl₂, we need 2 * 0.0225 mol = 0.0450 mol of AgNO₃. We only have 0.0200 mol of AgNO₃.
   • Therefore, AgNO₃ is the limiting reagent.

5. Calculate the Theoretical Yield:

   • From the balanced equation, 2 mol of AgNO₃ produces 2 mol of AgCl. Thus, the mole ratio is 1:1.
   • Moles of AgCl = 0.0200 mol (since AgNO₃ is the limiting reagent)
   • Mass of AgCl = 0.0200 mol x 143.32 g/mol = 2.8664 g
   • The mass of AgCl precipitate formed is approximately 2.87 g.

Problem 3:

Consider the reaction between zinc and hydrochloric acid:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

If 5.0 g of zinc are reacted with 100.0 mL of 1.0 M hydrochloric acid, determine the limiting reagent and the volume of hydrogen gas produced at standard temperature and pressure (STP).

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of Zn = 5.0 g / 65.38 g/mol = 0.0765 mol
   • Moles of HCl = (0.100 L) x (1.0 mol/L) = 0.100 mol

3. Determine the Mole Ratio:

   • Mole ratio of Zn to HCl = 0.0765 mol / 0.100 mol = 0.765

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of Zn to HCl is 1:2.
   • To consume 0.0765 mol of Zn, we need 2 * 0.0765 mol = 0.153 mol of HCl. We only have 0.100 mol of HCl.
   • Therefore, HCl is the limiting reagent.

5. Calculate the Theoretical Yield of H₂:

   • From the balanced equation, 2 mol of HCl produces 1 mol of H₂.
   • Moles of H₂ = 0.100 mol HCl / 2 = 0.050 mol H₂
   • At STP, 1 mole of any gas occupies 22.4 L.
   • Volume of H₂ = 0.050 mol x 22.4 L/mol = 1.12 L
   • The volume of hydrogen gas produced at STP is 1.12 L.

Problem 4:

Iron(III) oxide reacts with carbon monoxide to produce iron metal and carbon dioxide:

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

If 160.0 g of Fe₂O₃ is reacted with 84.0 g of CO, determine the limiting reagent and the mass of iron metal produced.

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of Fe₂O₃ = 160.0 g / 159.69 g/mol = 1.002 mol ≈ 1 mol
   • Moles of CO = 84.0 g / 28.01 g/mol = 2.999 mol ≈ 3 mol

3. Determine the Mole Ratio:

   • Mole ratio of Fe₂O₃ to CO = 1 mol / 3 mol = 1:3

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of Fe₂O₃ to CO is 1:3.
   • We have 1 mole of Fe₂O₃ and 3 moles of CO, which perfectly matches the stoichiometric ratio. Therefore, neither reactant is in excess.
   • However, because we rounded in the previous steps, it is important to check more closely.
     • To consume 1.002 mol of Fe₂O₃, we need 3 * 1.002 = 3.006 mol of CO. We only have 2.999 mol of CO, so CO is the limiting reagent.

5. Calculate the Theoretical Yield of Fe:

   • From the balanced equation, 3 mol of CO produces 2 mol of Fe.
   • Moles of Fe = (2/3) * 2.999 mol = 1.999 mol ≈ 2 mol
   • Mass of Fe = 1.999 mol x 55.845 g/mol = 111.63 g
   • The mass of iron metal produced is approximately 111.63 g.

Problem 5:

Consider the following reaction:

2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂

If 50.0 g of KMnO₄ and 75.0 g of HCl are used, which is the limiting reagent and how many grams of Cl₂ will be produced?

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of KMnO₄ = 50.0 g / 158.03 g/mol = 0.316 mol
   • Moles of HCl = 75.0 g / 36.46 g/mol = 2.057 mol

3. Determine the Mole Ratio:

   • Mole ratio of KMnO₄ to HCl = 0.316 mol / 2.057 mol = 0.154

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of KMnO₄ to HCl is 2:16, which simplifies to 1:8.
   • To consume 0.316 mol of KMnO₄, we need 8 * 0.316 mol = 2.528 mol of HCl. We only have 2.057 mol of HCl.
   • Therefore, HCl is the limiting reagent.

5. Calculate the Theoretical Yield of Cl₂:

   • From the balanced equation, 16 mol of HCl produces 5 mol of Cl₂.
   • Moles of Cl₂ = (5/16) * 2.057 mol = 0.643 mol
   • Mass of Cl₂ = 0.643 mol x 70.90 g/mol = 45.59 g
   • The mass of Cl₂ produced is approximately 45.59 g.

Problem 6:

In the reaction:

3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)

What mass of nitric acid (HNO₃) can be produced from the reaction of 42.0 g of NO₂ with 18.0 g of water?

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of NO₂ = 42.0 g / 46.01 g/mol = 0.913 mol
   • Moles of H₂O = 18.0 g / 18.02 g/mol = 0.999 mol ≈ 1 mol

3. Determine the Mole Ratio:

   • Mole ratio of NO₂ to H₂O = 0.913 mol / 0.999 mol = 0.914

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of NO₂ to H₂O is 3:1.
   • To consume 0.999 mol of H₂O, we need 3 * 0.999 mol = 2.997 mol of NO₂. We only have 0.913 mol of NO₂.
   • Therefore, NO₂ is the limiting reagent.

5. Calculate the Theoretical Yield of HNO₃:

   • From the balanced equation, 3 mol of NO₂ produces 2 mol of HNO₃.
   • Moles of HNO₃ = (2/3) * 0.913 mol = 0.609 mol
   • Mass of HNO₃ = 0.609 mol x 63.01 g/mol = 38.37 g
   • The mass of HNO₃ produced is approximately 38.37 g.

Problem 7:

Consider the combustion of methane (CH₄) with oxygen (O₂):

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

If 16.0 g of CH₄ is mixed with 48.0 g of O₂, identify the limiting reagent and determine the mass of CO₂ produced.

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of CH₄ = 16.0 g / 16.04 g/mol = 0.998 mol ≈ 1 mol
   • Moles of O₂ = 48.0 g / 32.00 g/mol = 1.5 mol

3. Determine the Mole Ratio:

   • Mole ratio of CH₄ to O₂ = 0.998 mol / 1.5 mol = 0.665

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of CH₄ to O₂ is 1:2.
   • To consume 0.998 mol of CH₄, we need 2 * 0.998 mol = 1.996 mol of O₂. We only have 1.5 mol of O₂.
   • Therefore, O₂ is the limiting reagent.

5. Calculate the Theoretical Yield of CO₂:

   • From the balanced equation, 2 mol of O₂ produces 1 mol of CO₂.
   • Moles of CO₂ = (1/2) * 1.5 mol = 0.75 mol
   • Mass of CO₂ = 0.75 mol x 44.01 g/mol = 33.01 g
   • The mass of CO₂ produced is approximately 33.01 g.

Problem 8:

For the reaction:

N₂O₄(g) → 2NO₂(g)

If 20.0 g of N₂O₄ is allowed to react, what mass of NO₂ will be produced, assuming the reaction goes to completion?

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of N₂O₄ = 20.0 g / 92.01 g/mol = 0.217 mol

3. Identify the Limiting Reagent:

   • In this case, there is only one reactant, so it is, by definition, the limiting reagent.

4. Calculate the Theoretical Yield of NO₂:

   • From the balanced equation, 1 mol of N₂O₄ produces 2 mol of NO₂.
   • Moles of NO₂ = 2 * 0.217 mol = 0.434 mol
   • Mass of NO₂ = 0.434 mol x 46.01 g/mol = 19.97 g
   • The mass of NO₂ produced is approximately 19.97 g.

Problem 9:

Consider the reaction:

SiCl₄(l) + 2Mg(s) → 2MgCl₂(s) + Si(s)

If 50.0 g of SiCl₄ is mixed with 25.0 g of Mg, determine the limiting reagent and the mass of silicon (Si) produced.

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of SiCl₄ = 50.0 g / 169.90 g/mol = 0.294 mol
   • Moles of Mg = 25.0 g / 24.31 g/mol = 1.028 mol

3. Determine the Mole Ratio:

   • Mole ratio of SiCl₄ to Mg = 0.294 mol / 1.028 mol = 0.286

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of SiCl₄ to Mg is 1:2.
   • To consume 0.294 mol of SiCl₄, we need 2 * 0.294 mol = 0.588 mol of Mg. We have 1.028 mol of Mg, so Mg is in excess.
   • Therefore, SiCl₄ is the limiting reagent.

5. Calculate the Theoretical Yield of Si:

   • From the balanced equation, 1 mol of SiCl₄ produces 1 mol of Si.
   • Moles of Si = 0.294 mol
   • Mass of Si = 0.294 mol x 28.09 g/mol = 8.26 g
   • The mass of silicon produced is approximately 8.26 g.

Problem 10:

What mass of carbon dioxide is produced when 20.0 g of ethyne (C₂H₂) is burned in 50.0 g of oxygen?

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

Solution:

1. Balanced Chemical Equation: The equation is already balanced.

2. Convert Given Information to Moles:

   • Moles of C₂H₂ = 20.0 g / 26.04 g/mol = 0.768 mol
   • Moles of O₂ = 50.0 g / 32.00 g/mol = 1.563 mol

3. Determine the Mole Ratio:

   • Mole ratio of C₂H₂ to O₂ = 0.768 mol / 1.563 mol = 0.491

4. Identify the Limiting Reagent:

   • From the balanced equation, the stoichiometric ratio of C₂H₂ to O₂ is 2:5, which is 0.4.
   • To consume 0.768 mol of C₂H₂, we need (5/2)*0.768 = 1.92 mol of O₂. We only have 1.563 mol of O₂.
   • Therefore, O₂ is the limiting reagent.

5. Calculate the Theoretical Yield of CO₂:

   • From the balanced equation, 5 mol of O₂ produces 4 mol of CO₂.
   • Moles of CO₂ = (4/5) * 1.563 mol = 1.250 mol
   • Mass of CO₂ = 1.250 mol * 44.01 g/mol = 55.01 g
   • The mass of carbon dioxide produced is approximately 55.01 g.

FAQ on Limiting Reagents

• Q: Why is it important to identify the limiting reagent?

  • A: Identifying the limiting reagent is crucial for determining the maximum amount of product that can be formed in a reaction (theoretical yield). It allows for accurate stoichiometric calculations and predictions.

• Q: What happens to the excess reagent(s)?

  • A: The excess reagent(s) are the reactants that are present in a greater amount than required to react completely with the limiting reagent. Some of the excess reagent will remain unreacted after the reaction is complete.

• Q: How does the limiting reagent affect the percent yield?

  • A: The percent yield is calculated by dividing the actual yield of the product by the theoretical yield (which is determined by the limiting reagent) and multiplying by 100%. The limiting reagent directly influences the theoretical yield, which in turn affects the percent yield.

• Q: Can a reaction have more than one limiting reagent?

  • A: No, a reaction can only have one limiting reagent. The limiting reagent is the reactant that is completely consumed first, thereby limiting the amount of product that can be formed.

• Q: Is the limiting reagent always the reactant with the smallest mass?

  • A: No, the limiting reagent is not necessarily the reactant with the smallest mass. It depends on the stoichiometry of the reaction and the molar masses of the reactants. You must convert masses to moles and compare the mole ratios to determine the limiting reagent.

Conclusion

Mastering the concept of the limiting reagent is fundamental to success in chemistry. By understanding the steps involved in identifying the limiting reagent and calculating theoretical yields, you can confidently tackle a wide range of stoichiometric problems. The practice problems provided in this article serve as a valuable resource for solidifying your understanding and honing your problem-solving skills. Remember to always balance the chemical equation, convert masses to moles, determine the mole ratio, and carefully compare it to the stoichiometric ratio. With practice and dedication, you'll become proficient in navigating the complexities of limiting reagent calculations and achieving accurate results.