Formula
Mole-mass-particle relations
Number of moles n = (mass)/(molar mass M) = (number of particles)/(NA) = (volume of gas at STP in L)/22.4. STP: 273.15 K, 100 kPa.
-- NCERT Class 11 Chemistry, Ch. 1, p. 12Stoichiometry
8 MCQs6 revision cards9-step worked example
Source: NCERT Some Basic Concepts of ChemistryPYQ coverage: NEET 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap that costs marks in stoichiometry is deceptively simple: students identify the limiting reagent by comparing absolute moles instead of dividing each reactant's moles by its stoichiometric coefficient. The reagent with the smaller ratio limits the reaction — not the one with fewer moles.
Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation. A balanced equation conserves atoms: the coefficients tell you the mole ratio in which substances react and form (NCERT Class 11 Chemistry Chapter 1, page 18).
Three routes to moles underpin every stoichiometric calculation:
- From mass: n = mass / molar mass
- From particles: n = N / N_A (6.022 × 10²³ mol⁻¹)
- From gas volume at STP: n = V / 22.4 L
Once moles are known, the balanced equation's coefficients scale reactants to products.
Limiting reagent identification: Given moles of two reactants, compute (moles available ÷ coefficient) for each. The smaller value identifies the limiting reagent. All product calculations use the limiting reagent's moles.
Percentage purity trap: When a sample is stated as X% pure, the effective mass available for reaction is sample mass × (X/100). Forgetting this step inflates the calculated moles and yields a wrong answer — a common distractor in NEET options.
Molarity and molality connect stoichiometry to solution chemistry: M = n/V (volume of solution in litres); m = n/mass of solvent in kg. Note: molarity uses solution volume; molality uses solvent mass.
Watch-out: 22.4 L/mol applies strictly at STP (273.15 K, 1 bar). If the question specifies non-STP conditions, use PV = nRT instead.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
What is the molar volume of an ideal gas at STP (273.15 K, 1 bar)?
MCQ 2Easy RecallPractice
Molality is defined as:
MCQ 3Easy RecallPractice
In a balanced chemical equation, stoichiometric coefficients represent:
MCQ 4Direct ApplicationPractice
For the reaction N₂ + 3H₂ → 2NH₃, if 2 mol N₂ and 4 mol H₂ are mixed, identify the limiting reagent.
MCQ 5Direct ApplicationPractice
A 50 g sample of CaCO₃ is 80% pure. How many moles of pure CaCO₃ are available for reaction? (Molar mass CaCO₃ = 100 g/mol)
MCQ 6Direct ApplicationPractice
5.0 g of NaOH (molar mass = 40 g/mol) is dissolved in water to make 250 mL of solution. What is the molarity?
MCQ 7CalculationPractice
For the reaction 2Al + 6HCl → 2AlCl₃ + 3H₂, if 5.4 g Al (molar mass 27) reacts with 300 mL of 1.0 M HCl, what is the maximum volume of H₂ produced at STP?
MCQ 8CalculationPractice
A 10.0 g impure sample of MgCO₃ (molar mass 84, 75% pure) is heated: MgCO₃ → MgO + CO₂. What mass of CO₂ is released? (Molar mass CO₂ = 44 g/mol)
Quick recall before you leave
Worked Example
- 1
Given
- Sample mass = 25.0 g - Purity = 80.0% - Molar mass CaCO₃ = 100 g/mol (exact, defined) - Molar volume at STP = 22.4 L/mol
- 2
Required
Volume of CO₂ at STP.
- 3
Concept
Percentage purity → effective mass → moles via molar mass → stoichiometric ratio → gas volume at STP.
- 4
Formula
Effective mass = sample mass × (purity/100) n = effective mass / M V = n × 22.4 (STP)
- 5
Substitution
Effective mass = 25.0 × (80.0/100) = 20.0 g n(CaCO₃) = 20.0 / 100 = 0.200 mol From 1:1 stoichiometry: n(CO₂) = 0.200 mol V(CO₂) = 0.200 × 22.4
- 6
Calculation
V = 4.48 L
- 7
Final answer
Volume of CO₂ at STP = 4.48 L Note on exact values: the molar mass 100 g/mol for CaCO₃ and the molar volume 22.4 L/mol are treated as exact defined constants in this context and do not limit significant figures. The answer is reported to 3 significant figures, matching the given mass (25.0 g) and purity (80.0%).
- 8
Common trap
Forgetting the purity step gives 25.0/100 × 22.4 = 5.60 L — a common wrong option that inflates the answer by 25%.
- 9
Similar NEET-style question
A 12.0 g sample of impure MgCO₃ (90% pure, M = 84 g/mol) decomposes on heating. Calculate the volume of CO₂ at STP. (Answer: 2.88 L) ---
Before solving, remember these
Definition
Limiting reagent
Reactant that is completely consumed first in a reaction; limits the amount of product. Excess reagent is the other reactant.
-- NCERT Class 11 Chemistry, Ch. 1, p. 18Formula
Concentration in solutions
Molarity (M) = moles of solute / volume of solution in L. Molality (m) = moles of solute / kg of solvent. Mole fraction x_A = n_A / (n_A + n_B + ...).
-- NCERT Class 11 Chemistry, Ch. 1, p. 20Formulas
Molality
Molal concentration: moles of solute per kg of solvent. Temperature-independent.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m | molality | mol/kg |
| n | moles solute | mol |
Valid when
- Mass of SOLVENT (not solution)
Molarity
Molar concentration: moles of solute per litre of solution.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | molarity | mol/L |
| n | moles solute | mol |
| V | solution volume | L |
Valid when
- Volume of SOLUTION not solvent
- Temperature dependent (volume changes with T)
Number of moles
Three equivalent ways to compute moles: from mass and molar mass, from number of particles, from gas volume at STP.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | moles | mol |
| m | mass | g |
| M | molar mass | g/mol |
| N | particle count | - |
| NA | Avogadro 6.022e23 | 1/mol |
| V | gas volume | L |
Valid when
- Use g/(g/mol) for mass route
- STP for gas volume route
- Pure substance
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student picks reagent with smaller absolute moles as limiting, ignoring stoichiometric coefficients.
When it triggers
Stoichiometry problem with two reactants.
How to avoid
Compute (moles available / coefficient) for each. Smaller value = limiting reagent. Compare ratios, not raw moles.
Category: Overthinking
Student computes from total mole amounts without checking stoichiometric ratios. Limiting reagent is the one that runs out first; rest is excess.
When it triggers
Stoichiometry problem with two reactants given.
How to avoid
Compare moles available to stoichiometric requirement. Compute moles_actual / coefficient for each; smaller value = limiting reagent.
Category: Unit Conversion
Student uses 22.4 L/mol at non-STP conditions or assumes ideal-gas-only molar volume.
When it triggers
Question gives gas at non-STP T or P.
How to avoid
22.4 L/mol applies ONLY at STP (273.15 K, 100 kPa). For other conditions, use PV=nRT.
Category: Overthinking
Student forgets to multiply by purity% before computing moles from impure sample mass.
When it triggers
Question states sample is X% pure (e.g. limestone, ore).
How to avoid
Effective mass = sample mass × (purity/100). Then compute moles using effective mass / molar mass.
Root cause: concept gap
Correction
Compute (moles_actual / coefficient) for each reactant. Smaller value = limiting.
Root cause: concept gap
Correction
22.4 L/mol applies ONLY at STP. Use PV=nRT for other conditions.
Root cause: concept gap
Correction
Effective mass = stated mass × (purity/100). Then convert to moles using molar mass.
Past Year Questions
4 questions from NEET 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
Dalton’s Atomic theory could not explain which of the following?
1Law of gaseous volume
2Law of conservation of mass
3Law of constant proportion
4Law of multiple proportion
NTA Answer: Option 1(final)
NEET 2024Revised key
1750 mg
2250 mg
3Zero mg
4200 mg
NTA Answer: Option 2(revised_final)
NEET 2024Revised key
1A BC 2 2
2ABC 3
3AB C 2 2
4ABC 4
NTA Answer: Option 2(revised_final)
NEET 2023
11.76 g
22.64 g
31.32 g
41.12 g
NTA Answer: Option 1(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Mole concept and stoichiometry calculations: mole-mass conversion, limiting reagent, percentage purity, % composition.
Direct ApplicationMedium
Common distractors
swapped classes
Tempts surface-level recall.
Sources
NCERT refs: Class 11 Chemistry Chapter 1, p.18
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