Limiting Reagent Practice Problems With Answers

Limiting Reagent Practice Problems with Answers
Master the concept of the limiting reagent through hands‑on examples and detailed solutions.

Introduction

In stoichiometry, the limiting reagent is the reactant that is completely consumed first, thereby determining the maximum amount of product that can form. Understanding how to identify the limiting reagent is essential for predicting yields, planning experiments, and scaling reactions safely. This article presents a variety of practice problems—ranging from simple single‑step reactions to more complex multi‑reactant systems—along with step‑by‑step solutions. By working through these examples, you’ll reinforce the methodology for balancing equations, converting units, and applying mole ratios to real‑world scenarios.

How to Identify the Limiting Reagent: A Quick Recap

1. Balance the equation.
2. Convert all given amounts to moles (using molar masses or molarity × volume).
3. Calculate the mole ratio of each reactant to the product from the balanced equation.
4. Determine the theoretical yield for each reactant by dividing its mole amount by its stoichiometric coefficient.
5. The reactant with the smallest theoretical yield is the limiting reagent.

Practice Problems

Problem 1 – Simple Acid‑Base Neutralization

Reactants: 50 g of NaOH and 60 g of H₂SO₄ react to form sodium sulfate and water.
Question: Which is the limiting reagent, and what mass of Na₂SO₄ can be produced?

Solution

1. Balanced equation
   [ 2,\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2,\text{H}_2\text{O} ]
2. Molar masses
   NaOH = 40 g mol⁻¹, H₂SO₄ = 98 g mol⁻¹, Na₂SO₄ = 142 g mol⁻¹.
3. Convert to moles
   [ n_{\text{NaOH}} = \frac{50}{40} = 1.25\ \text{mol} ] [ n_{\text{H}_2\text{SO}_4} = \frac{60}{98} \approx 0.612\ \text{mol} ]
4. Theoretical yields
   • From NaOH: ( \frac{1.25}{2} = 0.625\ \text{mol Na}_2\text{SO}_4)
   • From H₂SO₄: (0.612\ \text{mol Na}_2\text{SO}_4)
     The smaller value comes from H₂SO₄, so it is the limiting reagent.
5. Mass of Na₂SO₄ produced
   (0.612\ \text{mol} \times 142\ \text{g mol}^{-1} \approx 86.9\ \text{g}).

Answer: H₂SO₄ is limiting; 86.9 g of Na₂SO₄ can be produced Which is the point..

Problem 2 – Redox Reaction in Solution

Reactants: 10 mL of 0.5 M Fe³⁺ solution and 5 mL of 0.2 M Cu²⁺ solution are mixed. The reaction is
[ 2,\text{Fe}^{3+} + 3,\text{Cu}^{2+} \rightarrow 2,\text{Fe}^{2+} + 3,\text{Cu}^{+} ]
Question: Identify the limiting reagent and calculate the volume of the resulting solution that would contain 0.1 M Fe²⁺.

Solution

1. Moles of each ion
   [ n_{\text{Fe}^{3+}} = 0.5\ \text{M} \times 0.010\ \text{L} = 0.005\ \text{mol} ] [ n_{\text{Cu}^{2+}} = 0.2\ \text{M} \times 0.005\ \text{L} = 0.001\ \text{mol} ]
2. Stoichiometric requirement
   For every 2 mol Fe³⁺, 3 mol Cu²⁺ are needed.
   • Fe³⁺ needed for 0.001 mol Cu²⁺: ( \frac{2}{3} \times 0.001 \approx 0.000667\ \text{mol}).
     Since we have 0.005 mol Fe³⁺, Cu²⁺ is limiting.
3. Fe²⁺ produced
   From the balanced equation, 2 mol Fe³⁺ produce 2 mol Fe²⁺, so the ratio is 1:1.
   Because of this, 0.001 mol Cu²⁺ consumes 0.001 mol Fe³⁺, leaving (0.005-0.001 = 0.004) mol Fe³⁺ unreacted, but this does not affect Fe²⁺ yield.
   Fe²⁺ produced = 0.001 mol (since 1 mol Fe³⁺ → 1 mol Fe²⁺).
4. Desired concentration
   (C = 0.1\ \text{M} = \frac{0.001\ \text{mol}}{V}) → (V = \frac{0.001}{0.1} = 0.01\ \text{L} = 10\ \text{mL}).

Answer: Cu²⁺ is limiting; a 10 mL volume of the final solution contains 0.1 M Fe²⁺.

Problem 3 – Combustion Reaction

Reactants: 3 g of C₆H₁₂O₆ (glucose) reacts with excess O₂.
Question: How many grams of CO₂ are produced?

Solution

1. Balanced equation
   [ \text{C}6\text{H}{12}\text{O}_6 + 6,\text{O}_2 \rightarrow 6,\text{CO}_2 + 6,\text{H}_2\text{O} ]
2. Molar mass of glucose = 180 g mol⁻¹.
   Moles of glucose = ( \frac{3}{180} = 0.0167\ \text{mol}).
3. CO₂ produced
   1 mol glucose → 6 mol CO₂.
   Moles CO₂ = (0.0167 \times 6 = 0.100\ \text{mol}).
   Mass CO₂ = (0.100\ \text{mol} \times 44\ \text{g mol}^{-1} = 4.4\ \text{g}).

Answer: 4.4 g of CO₂ are produced It's one of those things that adds up. Less friction, more output..

Problem 4 – Multi‑Step Reaction (Sequential Limiting)

Step 1: 0.2 mol of A reacts with excess B to form C.
Step 2: 0.1 mol of C then reacts with excess D to produce E.
Question: Which reactant ultimately limits the amount of E produced, and how many moles of E can be obtained?

Solution

1. Step 1 stoichiometry (assume 1:1 for simplicity)
   0.2 mol A → 0.2 mol C.
2. Step 2 stoichiometry (1:1)
   0.1 mol C available → 0.1 mol E.
   Since only 0.1 mol C can react in step 2, the excess of C from step 1 remains unused.
   Because of this, C (produced in step 1) is the limiting reagent for step 2, not the original A.

Answer: C limits the production of E; 0.1 mol of E can be obtained.

Problem 5 – Gas‑Phase Reaction with Partial Pressures

Reactants: A mixture of 2 atm of N₂ and 5 atm of H₂ in a closed container at 298 K reacts to form ammonia:
[ \text{N}_2 + 3,\text{H}_2 \rightarrow 2,\text{NH}_3 ]
Question: Which gas is limiting, and what will be the partial pressure of NH₃ after the reaction reaches completion?

Solution

1. Mole ratios from partial pressures
   For ideal gases, partial pressures are proportional to mole numbers.
   • Required ratio: 1 atm N₂ : 3 atm H₂.
   • Actual ratio: 2 atm N₂ : 5 atm H₂ → 2 : 5 ≈ 1 : 2.5.
     H₂ is in excess; N₂ is limiting.
2. Moles of NH₃ produced
   1 mol N₂ → 2 mol NH₃.
   Moles N₂ = 2 atm (relative).
   Moles NH₃ = (2 \times 2 = 4) (relative units).
3. Total pressure after reaction
   Initial total = 2 + 5 = 7 atm.
   NH₃ replaces consumed gases:
   • N₂ consumed: 2 atm → 0 atm.
   • H₂ consumed: 3 × 2 = 6 atm → 5 – 6 = –1 atm?
     Wait, we must adjust: For every 1 atm N₂, 3 atm H₂ are needed.
     With 2 atm N₂, we need 6 atm H₂, but only 5 atm are present, so actually H₂ is limiting!
     Recalculate:
   • H₂ limiting: 5 atm H₂ → requires 5/3 ≈ 1.667 atm N₂.
   • N₂ available: 2 atm, so H₂ is limiting.
4. NH₃ produced
   3 atm H₂ → 2 atm NH₃.
   Partial pressure of NH₃ = 2 atm.

Answer: H₂ is the limiting reagent; the partial pressure of NH₃ after completion is 2 atm That's the part that actually makes a difference..

FAQ

Q1: Can the limiting reagent change during a reaction?
A1: In a single‑step reaction, the limiting reagent is fixed by initial amounts. Still, in multi‑step reactions or when intermediates are involved, the limiting reagent for a later step may differ from the first step’s limiting reagent.

Q2: What if the stoichiometric coefficients are not whole numbers?
A2: Multiply the entire balanced equation by the least common multiple of the denominators to obtain integer coefficients. This simplifies mole‑ratio calculations.

Q3: How does temperature affect the limiting reagent?
A3: Temperature can influence reaction rates but not the stoichiometric limiting reagent, assuming the reactants are fully mixed and no side reactions consume them preferentially Simple, but easy to overlook..

Conclusion

Mastering limiting reagent calculations equips you with a powerful tool for predicting reaction outcomes, optimizing resource usage, and scaling processes safely. By consistently applying the steps—balancing, converting to moles, comparing theoretical yields—you can confidently solve problems ranging from simple neutralizations to complex gas‑phase equilibria. Practice with diverse examples, and soon identifying the limiting reagent will become second nature That's the part that actually makes a difference..