Limiting Reactant Practice Problems And Answers

Delving into the world of chemical reactions often brings us face-to-face with the concept of the limiting reactant, the unsung hero (or villain, depending on how you look at it) that dictates the extent to which a reaction can proceed. Mastering this concept is crucial for anyone studying chemistry, whether you're a high school student or a seasoned researcher. Let's unravel the mysteries of limiting reactants with a series of practice problems and detailed explanations.

Understanding the Limiting Reactant

Before we dive into the problems, let's solidify our understanding of what a limiting reactant actually is.

In a chemical reaction, reactants combine in specific stoichiometric ratios to form products. However, in real-world scenarios, we rarely have reactants present in those exact ratios. One reactant will usually be consumed completely before the others. This reactant, the one that gets used up first, is the limiting reactant.

The limiting reactant determines the maximum amount of product that can be formed. The other reactants are present in excess, meaning there's more than enough of them to react with the limiting reactant. They're left over after the reaction is complete.

Think of it like making sandwiches. If you have 10 slices of bread and 7 slices of cheese, you can only make 3 complete cheese sandwiches (assuming two slices of bread and one slice of cheese per sandwich). The cheese is the "limiting reactant" because you'll run out of it before you run out of bread. The bread is in excess.

Key Steps to Solving Limiting Reactant Problems

To successfully tackle limiting reactant problems, follow these steps:

1. Write a balanced chemical equation: This is the foundation. You cannot accurately solve the problem without a correctly balanced equation.

2. Convert reactant amounts to moles: Moles are the chemist's counting unit. Use molar mass to convert grams to moles, or use the ideal gas law (PV = nRT) to convert pressure and volume to moles, if applicable.

3. Determine the mole ratio: Use the balanced equation to find the stoichiometric ratio between the reactants and the desired product.

4. Identify the limiting reactant: Calculate how much product each reactant could form, assuming the other reactant is present in excess. The reactant that produces the least amount of product is the limiting reactant.

5. Calculate the theoretical yield: The amount of product formed based on the limiting reactant is the theoretical yield. This is the maximum possible amount of product you can obtain.

6. Calculate the percent yield (if required): If the problem provides the actual yield (the amount of product actually obtained in an experiment), you can calculate the percent yield using the following formula:

   Percent Yield = (Actual Yield / Theoretical Yield) * 100%

Limiting Reactant Practice Problems and Answers

Let's put these steps into practice with a series of problems:

Problem 1:

If 10.0 g of aluminum (Al) reacts with 35.0 g of chlorine gas (Cl₂), how many grams of aluminum chloride (AlCl₃) can be produced?

Answer:

1. Balanced Chemical Equation:

   2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

2. Convert Reactant Amounts to Moles:

   • Moles of Al = 10.0 g / 26.98 g/mol = 0.371 mol
   • Moles of Cl₂ = 35.0 g / 70.90 g/mol = 0.494 mol

3. Determine the Mole Ratio:

   From the balanced equation, 2 moles of Al react with 3 moles of Cl₂ to produce 2 moles of AlCl₃.

4. Identify the Limiting Reactant:

   • Based on Al: 0.371 mol Al can produce (0.371 mol Al) * (2 mol AlCl₃ / 2 mol Al) = 0.371 mol AlCl₃
   • Based on Cl₂: 0.494 mol Cl₂ can produce (0.494 mol Cl₂) * (2 mol AlCl₃ / 3 mol Cl₂) = 0.329 mol AlCl₃

   Since Cl₂ produces less AlCl₃, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of AlCl₃ is 0.329 mol. Convert this to grams:

   Grams of AlCl₃ = 0.329 mol * 133.34 g/mol = 43.9 g

6. Answer: 43.9 grams of aluminum chloride can be produced.

Problem 2:

Ammonia (NH₃) reacts with oxygen (O₂) to form nitrogen monoxide (NO) and water (H₂O). If 4.00 g of ammonia and 6.00 g of oxygen react, what is the limiting reactant and how many grams of nitrogen monoxide are formed?

Answer:

1. Balanced Chemical Equation:

   4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

2. Convert Reactant Amounts to Moles:

   • Moles of NH₃ = 4.00 g / 17.03 g/mol = 0.235 mol
   • Moles of O₂ = 6.00 g / 32.00 g/mol = 0.188 mol

3. Determine the Mole Ratio:

   From the balanced equation, 4 moles of NH₃ react with 5 moles of O₂ to produce 4 moles of NO.

4. Identify the Limiting Reactant:

   • Based on NH₃: 0.235 mol NH₃ can produce (0.235 mol NH₃) * (4 mol NO / 4 mol NH₃) = 0.235 mol NO
   • Based on O₂: 0.188 mol O₂ can produce (0.188 mol O₂) * (4 mol NO / 5 mol O₂) = 0.150 mol NO

   Since O₂ produces less NO, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of NO is 0.150 mol. Convert this to grams:

   Grams of NO = 0.150 mol * 30.01 g/mol = 4.50 g

6. Answer: The limiting reactant is oxygen (O₂), and 4.50 grams of nitrogen monoxide (NO) are formed.

Problem 3:

Consider the reaction between iron(III) oxide (Fe₂O₃) and carbon monoxide (CO) to produce iron (Fe) and carbon dioxide (CO₂). If 25.0 grams of Fe₂O₃ is reacted with 30.0 grams of CO, what is the limiting reactant, and what mass of iron is produced?

Answer:

1. Balanced Chemical Equation:

   Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

2. Convert Reactant Amounts to Moles:

   • Moles of Fe₂O₃ = 25.0 g / 159.69 g/mol = 0.157 mol
   • Moles of CO = 30.0 g / 28.01 g/mol = 1.07 mol

3. Determine the Mole Ratio:

   From the balanced equation, 1 mole of Fe₂O₃ reacts with 3 moles of CO to produce 2 moles of Fe.

4. Identify the Limiting Reactant:

   • Based on Fe₂O₃: 0.157 mol Fe₂O₃ can produce (0.157 mol Fe₂O₃) * (2 mol Fe / 1 mol Fe₂O₃) = 0.314 mol Fe
   • Based on CO: 1.07 mol CO can produce (1.07 mol CO) * (2 mol Fe / 3 mol CO) = 0.713 mol Fe

   Since Fe₂O₃ produces less Fe, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of Fe is 0.314 mol. Convert this to grams:

   Grams of Fe = 0.314 mol * 55.85 g/mol = 17.5 g

6. Answer: The limiting reactant is iron(III) oxide (Fe₂O₃), and 17.5 grams of iron (Fe) are produced.

Problem 4:

In a laboratory, 5.0 g of hydrogen gas (H₂) is reacted with 10.0 g of nitrogen gas (N₂). What is the theoretical yield of ammonia (NH₃) in grams?

Answer:

1. Balanced Chemical Equation:

   N₂(g) + 3H₂(g) → 2NH₃(g)

2. Convert Reactant Amounts to Moles:

   • Moles of H₂ = 5.0 g / 2.02 g/mol = 2.48 mol
   • Moles of N₂ = 10.0 g / 28.02 g/mol = 0.357 mol

3. Determine the Mole Ratio:

   From the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

4. Identify the Limiting Reactant:

   • Based on N₂: 0.357 mol N₂ can produce (0.357 mol N₂) * (2 mol NH₃ / 1 mol N₂) = 0.714 mol NH₃
   • Based on H₂: 2.48 mol H₂ can produce (2.48 mol H₂) * (2 mol NH₃ / 3 mol H₂) = 1.65 mol NH₃

   Since N₂ produces less NH₃, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of NH₃ is 0.714 mol. Convert this to grams:

   Grams of NH₃ = 0.714 mol * 17.03 g/mol = 12.2 g

6. Answer: The theoretical yield of ammonia (NH₃) is 12.2 grams.

Problem 5:

Methane (CH₄) reacts with water (H₂O) to produce hydrogen gas (H₂) and carbon monoxide (CO). If 100.0 g of methane reacts with 100.0 g of water, what mass of hydrogen gas is produced?

Answer:

1. Balanced Chemical Equation:

   CH₄(g) + H₂O(g) → 3H₂(g) + CO(g)

2. Convert Reactant Amounts to Moles:

   • Moles of CH₄ = 100.0 g / 16.04 g/mol = 6.23 mol
   • Moles of H₂O = 100.0 g / 18.02 g/mol = 5.55 mol

3. Determine the Mole Ratio:

   From the balanced equation, 1 mole of CH₄ reacts with 1 mole of H₂O to produce 3 moles of H₂.

4. Identify the Limiting Reactant:

   • Based on CH₄: 6.23 mol CH₄ can produce (6.23 mol CH₄) * (3 mol H₂ / 1 mol CH₄) = 18.7 mol H₂
   • Based on H₂O: 5.55 mol H₂O can produce (5.55 mol H₂O) * (3 mol H₂ / 1 mol H₂O) = 16.7 mol H₂

   Since H₂O produces less H₂, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of H₂ is 16.7 mol. Convert this to grams:

   Grams of H₂ = 16.7 mol * 2.02 g/mol = 33.7 g

6. Answer: 33.7 grams of hydrogen gas are produced.

Problem 6:

What mass of precipitate (PbCl₂) will be formed when 20.0 mL of 0.100 M Pb(NO₃)₂ reacts with 50.0 mL of 0.300 M NaCl?

Answer:

1. Balanced Chemical Equation:

   Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq)

2. Convert Reactant Amounts to Moles:

   • Moles of Pb(NO₃)₂ = (0.0200 L) * (0.100 mol/L) = 0.00200 mol
   • Moles of NaCl = (0.0500 L) * (0.300 mol/L) = 0.0150 mol

3. Determine the Mole Ratio:

   From the balanced equation, 1 mole of Pb(NO₃)₂ reacts with 2 moles of NaCl to produce 1 mole of PbCl₂.

4. Identify the Limiting Reactant:

   • Based on Pb(NO₃)₂: 0.00200 mol Pb(NO₃)₂ can produce (0.00200 mol Pb(NO₃)₂) * (1 mol PbCl₂ / 1 mol Pb(NO₃)₂) = 0.00200 mol PbCl₂
   • Based on NaCl: 0.0150 mol NaCl can produce (0.0150 mol NaCl) * (1 mol PbCl₂ / 2 mol NaCl) = 0.00750 mol PbCl₂

   Since Pb(NO₃)₂ produces less PbCl₂, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of PbCl₂ is 0.00200 mol. Convert this to grams:

   Grams of PbCl₂ = 0.00200 mol * 278.10 g/mol = 0.556 g

6. Answer: 0.556 grams of precipitate (PbCl₂) will be formed.

Problem 7:

If 5.0 g of zinc metal is added to 10.0 mL of 1.0 M hydrochloric acid, how many grams of hydrogen gas are produced?

Answer:

1. Balanced Chemical Equation:

   Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

2. Convert Reactant Amounts to Moles:

   • Moles of Zn = 5.0 g / 65.38 g/mol = 0.0765 mol
   • Moles of HCl = (0.0100 L) * (1.0 mol/L) = 0.0100 mol

3. Determine the Mole Ratio:

   From the balanced equation, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H₂.

4. Identify the Limiting Reactant:

   • Based on Zn: 0.0765 mol Zn can produce (0.0765 mol Zn) * (1 mol H₂ / 1 mol Zn) = 0.0765 mol H₂
   • Based on HCl: 0.0100 mol HCl can produce (0.0100 mol HCl) * (1 mol H₂ / 2 mol HCl) = 0.00500 mol H₂

   Since HCl produces less H₂, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of H₂ is 0.00500 mol. Convert this to grams:

   Grams of H₂ = 0.00500 mol * 2.02 g/mol = 0.0101 g

6. Answer: 0.0101 grams of hydrogen gas are produced.

Problem 8:

Consider the following reaction: C₆H₁₂O₆ (glucose) → 2C₂H₅OH (ethanol) + 2CO₂ (carbon dioxide). If you start with 50.0 grams of glucose, what is the theoretical yield of ethanol in grams?

Answer:

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

   C₆H₁₂O₆(s) → 2C₂H₅OH(l) + 2CO₂(g)

2. Convert Reactant Amounts to Moles:

   • Moles of C₆H₁₂O₆ = 50.0 g / 180.16 g/mol = 0.278 mol

3. Determine the Mole Ratio:

   From the balanced equation, 1 mole of C₆H₁₂O₆ produces 2 moles of C₂H₅OH. Since there's only one reactant, it's automatically the limiting reactant!

4. Identify the Limiting Reactant:

   Glucose is the limiting reactant.

5. Calculate the Theoretical Yield:

   Moles of ethanol produced = (0.278 mol glucose) * (2 mol ethanol / 1 mol glucose) = 0.556 mol ethanol Grams of ethanol = 0.556 mol * 46.07 g/mol = 25.6 g

6. Answer: The theoretical yield of ethanol is 25.6 grams.

Problem 9:

A reaction between 5.0 g of magnesium and excess oxygen yields 7.0 g of magnesium oxide. What is the percent yield of this reaction?

Answer:

1. Balanced Chemical Equation:

   2Mg(s) + O₂(g) → 2MgO(s)

2. Convert Reactant Amounts to Moles:

   *Moles of Mg = 5.0 g / 24.31 g/mol = 0.206 mol

3. Determine the Mole Ratio:

   From the balanced equation, 2 moles of Mg produce 2 moles of MgO. Since oxygen is in excess, magnesium is the limiting reactant.

4. Identify the Limiting Reactant:

   Magnesium is the limiting reactant.

5. Calculate the Theoretical Yield:

   Moles of MgO = 0.206 mol Grams of MgO = 0.206 mol * 40.30 g/mol = 8.30 g (theoretical yield)

6. Calculate Percent Yield:

   Percent Yield = (Actual Yield / Theoretical Yield) * 100% Percent Yield = (7.0 g / 8.30 g) * 100% = 84.3%

7. Answer: The percent yield of the reaction is 84.3%.

Problem 10:

If 20.0 g of silicon dioxide (SiO₂) is reacted with 60.0 g of hydrofluoric acid (HF) according to the following equation:

SiO₂(s) + 4HF(aq) → SiF₄(g) + 2H₂O(l)

What mass of silicon tetrafluoride (SiF₄) is produced?

Answer:

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

   SiO₂(s) + 4HF(aq) → SiF₄(g) + 2H₂O(l)

2. Convert Reactant Amounts to Moles:

   • Moles of SiO₂ = 20.0 g / 60.08 g/mol = 0.333 mol
   • Moles of HF = 60.0 g / 20.01 g/mol = 3.00 mol

3. Determine the Mole Ratio:

   From the balanced equation, 1 mole of SiO₂ reacts with 4 moles of HF to produce 1 mole of SiF₄.

4. Identify the Limiting Reactant:

   • Based on SiO₂: 0.333 mol SiO₂ can produce (0.333 mol SiO₂) * (1 mol SiF₄ / 1 mol SiO₂) = 0.333 mol SiF₄
   • Based on HF: 3.00 mol HF can produce (3.00 mol HF) * (1 mol SiF₄ / 4 mol HF) = 0.750 mol SiF₄

   Since SiO₂ produces less SiF₄, it is the limiting reactant.

5. Calculate the Theoretical Yield:

   The theoretical yield of SiF₄ is 0.333 mol. Convert this to grams:

   Grams of SiF₄ = 0.333 mol * 104.08 g/mol = 34.7 g

6. Answer: 34.7 grams of silicon tetrafluoride (SiF₄) are produced.

Common Mistakes to Avoid

• Forgetting to Balance the Equation: This is the most frequent mistake. An unbalanced equation leads to incorrect mole ratios and, consequently, wrong answers.
• Using Grams Directly in Mole Ratios: Mole ratios are based on moles, not grams. Always convert grams to moles before using the stoichiometric coefficients.
• Incorrectly Identifying the Limiting Reactant: Carefully compare the amount of product each reactant could produce. The smaller amount indicates the limiting reactant.
• Rounding Errors: Avoid rounding intermediate calculations. Round only the final answer to the appropriate number of significant figures.

Conclusion

Mastering limiting reactant problems is fundamental to understanding chemical reactions and stoichiometry. By following the outlined steps, carefully balancing equations, and converting to moles, you can confidently solve a wide range of problems. Remember to practice consistently and pay attention to common mistakes. With dedication, you'll become proficient in identifying limiting reactants and accurately predicting the yield of chemical reactions. Keep practicing, and you'll conquer those stoichiometry challenges in no time!