Stoichiometry Practice: 5 Limiting Reagent Questions with Answers
This guide presents five limiting reagent practice questions from the Dreamers Quiz bank with step-by-step solutions.
You will learn how to identify limiting reactants, calculate product yields, and solve numerical questions accurately under timed exam conditions.
Understanding stoichiometric ratios and limiting reactants
Stoichiometry measures quantitative relationships between reactants and products based on the law of conservation of mass. Because matter cannot be created or destroyed, atoms must balance across both sides of a reaction equation. The term comes from the Greek stoicheion (element) and metron (measure).
In exam problems, reactants are rarely mixed in exact proportions. The limiting reagent runs out first and halts product formation, leaving the excess reagent partially unreacted. Follow this calculation sequence:
- Balance the chemical equation.
- Convert given masses or volumes into moles.
- Divide reactant moles by stoichiometric coefficients.
- Identify the limiting reagent from the lowest ratio.
- Calculate theoretical yield using the limiting reagent.
| Term | Definition | Calculation role |
|---|---|---|
| Stoichiometric coefficient | Formula multiplier in a balanced equation. | Sets molar reaction ratios. |
| Limiting reagent | Reactant consumed completely first. | Determines maximum product yield. |
| Excess reagent | Reactant remaining unreacted after completion. | Subtract reacted from initial moles. |
| Theoretical yield | Maximum product predicted by stoichiometry. | Derived from limiting reagent moles. |
Limiting reagent sample questions: set 1, 2, and 3
These three questions from our question bank show typical problem formats tested in school and competitive entrance exams.
Question 1: Combustion of methane
Methane burns in oxygen: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). If 1.0 mol of CH4 reacts with 1.5 mol of O2, identify the limiting reagent and find the moles of CO2 produced.
Solution:
- Compare ratios: CH4: 1.0 / 1 = 1.0. O2: 1.5 / 2 = 0.75.
- Find limiting reactant: O2 gives the smaller quotient (0.75 < 1.0), so O2 is limiting.
- Calculate yield: 2 mol O2 yields 1 mol CO2. Moles of CO2 = 1.5 × (1 / 2) = 0.75 mol.
Answer: O2 is limiting; 0.75 mol of CO2 is produced.
Question 2: Synthesis of water from mass quantities
Hydrogen and oxygen form water: 2H2(g) + O2(g) → 2H2O(l). If 6.0 g of H2 reacts with 32.0 g of O2, identify the limiting reagent and the mass of water formed. (Molar masses: H = 1.0 g/mol, O = 16.0 g/mol).
Solution:
- Convert to moles: H2: 6.0 / 2.0 = 3.0 mol. O2: 32.0 / 32.0 = 1.0 mol.
- Compare ratios: H2: 3.0 / 2 = 1.5. O2: 1.0 / 1 = 1.0. O2 is limiting.
- Calculate mass: 1.0 mol O2 yields 2.0 mol H2O. Mass = 2.0 × 18.0 = 36.0 g. Leftover H2 = 3.0 - 2.0 = 1.0 mol.
Answer: O2 is limiting; 36.0 g of water is formed.
Question 3: Synthesis of hydrogen chloride gas
Hydrogen gas reacts with chlorine gas: H2(g) + Cl2(g) → 2HCl(g). If 4.0 g of H2 reacts with 71.0 g of Cl2, find the limiting reagent and calculate the mass of HCl produced. (Molar masses: H = 1.0 g/mol, Cl = 35.5 g/mol).
Solution:
- Convert to moles: H2: 4.0 / 2.0 = 2.0 mol. Cl2: 71.0 / 71.0 = 1.0 mol.
- Compare ratios: H2: 2.0 / 1 = 2.0. Cl2: 1.0 / 1 = 1.0. Cl2 is limiting.
- Calculate mass: 1.0 mol Cl2 yields 2.0 mol HCl. Mass = 2.0 × 36.5 = 73.0 g. Leftover H2 = 2.0 - 1.0 = 1.0 mol.
Answer: Cl2 is limiting; 73.0 g of HCl is formed.
Limiting reagent sample questions: set 4 and 5
These two questions address gas volume calculations and unreacted excess reagents as covered in NCERT textbooks.
Question 4: Metal and acid reaction
Magnesium reacts with hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g). If 24.3 g of Mg reacts with 36.5 g of HCl, find the limiting reagent and the volume of H2 gas produced at STP (22.4 L/mol). (Molar masses: Mg = 24.3 g/mol, HCl = 36.5 g/mol).
Solution:
- Convert to moles: Mg: 24.3 / 24.3 = 1.0 mol. HCl: 36.5 / 36.5 = 1.0 mol.
- Compare ratios: Mg: 1.0 / 1 = 1.0. HCl: 1.0 / 2 = 0.5. HCl is limiting.
- Calculate gas volume: 2 mol HCl yields 1 mol H2. Moles of H2 = 1.0 × (1 / 2) = 0.5 mol. Volume at STP = 0.5 × 22.4 = 11.2 L.
Answer: HCl is limiting; 11.2 L of H2 gas is formed at STP.
Question 5: Sulfur dioxide oxidation and excess reagent
Sulfur dioxide oxidizes according to: 2SO2(g) + O2(g) → 2SO3(g). A vessel holds 4.0 mol of SO2 and 3.0 mol of O2. Find the limiting reagent, moles of SO3 formed, and moles of excess reagent left.
Solution:
- Compare ratios: SO2: 4.0 / 2 = 2.0. O2: 3.0 / 1 = 3.0. SO2 is limiting.
- Calculate product: 2 mol SO2 yields 2 mol SO3, so 4.0 mol SO2 yields 4.0 mol SO3.
- Calculate excess: O2 used = 4.0 × (1 / 2) = 2.0 mol. Leftover O2 = 3.0 - 2.0 = 1.0 mol.
Answer: SO2 is limiting; 4.0 mol of SO3 forms and 1.0 mol of O2 remains.
How to practice speed and accuracy for timed exams
In competitive exams like NEET, CUET, and Class 11 tests, solving limiting reagent problems quickly requires discipline. Many students mistakenly compare raw masses instead of converting to moles, or forget to divide moles by balanced coefficients. Always balance your equation first and write down mole-to-coefficient ratios clearly.
To build speed, practice under clock pressure using the Dreamers Quiz portal. How it works: pick a class or exam, attempt the set against the clock, then see your score and the answer key. Attempt as many times as you like. Use it for chapter revision before a school test, a quick check after a class, or speed practice for the NEET and CUET MCQ sections.
Key takeaways
- Always start by writing a fully balanced chemical equation.
- Convert all given quantities into moles; never compare raw masses directly.
- Divide reactant moles by their balanced coefficients to identify the limiting reagent.
- Calculate product yields and excess reactant consumption using only the limiting reagent.
- Practice timed quizzes regularly to improve calculation speed for NEET, CUET, and school tests.
Frequently asked questions
Can a reactant present in a lower mass be the excess reagent?
Yes. Chemical reactions proceed according to mole ratios rather than physical masses. A reactant with a lower mass can supply more moles if its molar mass is low. Always convert mass to moles before determining which substance is consumed first.
What does a stoichiometric ratio represent in a reaction?
A stoichiometric ratio represents the optimum proportion where all reactants are consumed completely. Under this condition, assuming the reaction proceeds to completion, there is no deficiency and no excess of any reagent.
What fundamental law governs stoichiometry calculations?
Stoichiometry is based on the law of conservation of mass. Because matter cannot be created or destroyed during a chemical reaction, the total mass of reactants must equal the total mass of products, resulting in a balanced ratio of positive integers.
What is the difference between reaction stoichiometry and gas stoichiometry?
Reaction stoichiometry measures quantitative relationships among substances during chemical reactions. Gas stoichiometry deals specifically with reactions involving gases, assuming ideal gas behavior at known temperature, pressure, and volume.
What to do next. Try it yourself on the page below.
Chemistry Quiz — Free Timed Chemistry Quizzes →
