Let us consider the combustion of methane. A balanced equation for this reaction is as given below: CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) … The coefficients 2 for O2 and H2O are called stoichiometric coefficients. Similarly the coefficient for CH4 and CO2 is one in each case. They represent the number of molecules (and moles as well) taking part in the reaction or formed in the reaction. Thus, according to the above chemical reaction, • One mole of CH4(g) reacts with two moles of O2(g) to give one mole of CO2(g) and two moles of H2O(g) • One molecule of CH4(g) reacts with 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g) • 22.7 L of CH4(g) reacts with 45.4 L of O2 (g) to give 22.7 L of CO2 (g) and 45.4 L of H2O(g) • 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2×18 g of H2O (g). From these relationships, the given data can be interconverted as follows: mass ⇌ moles ⇌ no. of molecules Mass / Volume = Density
-- NCERT Class 11 Chemistry, Ch. 1, p. 20Stoichiometry
Stoichiometry, explained for NEET
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 20).
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.
Can you answer these Stoichiometry MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
What is the molar volume of an ideal gas at STP (273.15 K, 1 bar)?
Show answer and why every option is right or wrong
Answer: B. The molar volume of any ideal gas at STP is 22.4 L mol⁻¹ (NCERT Class 11 Chemistry Chapter 1, page 12).
Why A is wrong: A is wrong because 11.2 L is the volume of 0.5 mol of gas at STP, not 1 mol (trap: halving the molar volume).
Why C is wrong: C is wrong because 44.8 L corresponds to 2 mol of gas at STP, not 1 mol (trap: doubling error).
Why D is wrong: D is wrong because 6.022 × 10²³ is Avogadro's number (particles per mole), not a volume value (trap: confusing N_A with molar volume).
Molality is defined as:
Show answer and why every option is right or wrong
Answer: A. Molality (m) = moles of solute / mass of solvent in kg. It is temperature-independent because mass does not change with temperature (NCERT Class 11 Chemistry Chapter 1, page 20).
Why B is wrong: B is wrong because moles of solute per litre of solution defines molarity, not molality (trap: confusing molarity and molality).
Why C is wrong: C is wrong because the denominator must be mass of solvent, not mass of solution — including solute mass in the denominator gives an incorrect concentration value (trap: solvent vs solution confusion).
Why D is wrong: D is wrong because molality uses moles (not grams) of solute and kilograms (not litres) of solvent — this definition matches no standard concentration unit (trap: mixing mass and volume units).
In a balanced chemical equation, stoichiometric coefficients represent:
Show answer and why every option is right or wrong
Answer: B. Coefficients in a balanced equation give the mole ratio in which substances react and form (NCERT Class 11 Chemistry Chapter 1, page 20).
Why A is wrong: A is wrong because coefficients give mole ratios, not mass ratios — mass ratios require multiplying coefficients by respective molar masses (trap: conflating moles with mass).
Why C is wrong: C is wrong because the volume ratio equals the mole ratio ONLY for gases at the same T and P, not at any conditions and not for solids/liquids (trap: overgeneralising Avogadro's law).
Why D is wrong: D is wrong because coefficients represent the number of formula units (moles) of each substance, not the number of atoms within a single molecule (trap: confusing subscripts with coefficients).
For the reaction N₂ + 3H₂ → 2NH₃, if 2 mol N₂ and 4 mol H₂ are mixed, identify the limiting reagent.
Show answer and why every option is right or wrong
Answer: A. Compare moles/coefficient: N₂ → 2/1 = 2; H₂ → 4/3 = 1.33. The smaller ratio (H₂) is the limiting reagent (NCERT Class 11 Chemistry Chapter 1, page 20; trap: limiting reagent by stoichiometric ratio).
Why B is wrong: B is wrong because having fewer absolute moles does not determine the limiting reagent — you must divide by the stoichiometric coefficient. Here N₂ actually has the larger ratio (2/1 = 2) so it is in excess (trap: comparing raw moles instead of mole-to-coefficient ratios).
Why C is wrong: C is wrong because the reagent with the SMALLER ratio is limiting, not the larger. N₂'s ratio (2) > H₂'s ratio (1.33), so N₂ is in excess (trap: inverting the comparison logic).
Why D is wrong: D is wrong because 2 mol N₂ would require 6 mol H₂ by stoichiometry, but only 4 mol H₂ is available — H₂ runs out first, leaving unreacted N₂ (trap: assuming both reactants are consumed when amounts are given).
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)
Show answer and why every option is right or wrong
Answer: C. Effective mass = 50 × (80/100) = 40 g. Moles = 40/100 = 0.40 mol (trap: percentage purity — must apply purity before dividing by molar mass).
Why A is wrong: A is wrong because 0.50 mol comes from 50/100, which ignores the 80% purity — you used the total sample mass instead of the effective pure mass (trap: forgetting to apply purity percentage).
Why B is wrong: B is wrong because 0.80 is the purity fraction itself (80/100), not the mole value — you confused the decimal purity with the number of moles (trap: arithmetic shortcut error).
Why D is wrong: D is wrong because 0.625 comes from 50/(80), which incorrectly divides sample mass by the percentage number rather than first computing effective mass = 50 × 0.80 (trap: inverting the purity calculation).
5.0 g of NaOH (molar mass = 40 g/mol) is dissolved in water to make 250 mL of solution. What is the molarity?
Show answer and why every option is right or wrong
Answer: D. Moles NaOH = 5.0/40 = 0.125 mol. Volume = 250 mL = 0.250 L. M = 0.125/0.250 = 0.50 M (NCERT Class 11 Chemistry Chapter 1, page 20).
Why A is wrong: A is wrong because 0.020 comes from 5.0/250 = 0.020, which divides mass directly by volume in mL without first converting to moles (trap: skipping the mass-to-moles conversion).
Why B is wrong: B is wrong because 0.125 is the number of moles, not the molarity — you forgot to divide by the volume in litres (trap: reporting moles as molarity).
Why C is wrong: C is wrong because 2.0 M results from dividing moles by volume in mL (0.125/0.250 ≠ 2.0; this likely comes from inverting to 0.250/0.125) or from an arithmetic error (trap: inverting the n/V division).
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?
Show answer and why every option is right or wrong
Answer: D. Moles Al = 5.4/27 = 0.20 mol; ratio = 0.20/2 = 0.10. Moles HCl = 0.300 × 1.0 = 0.30 mol; ratio = 0.30/6 = 0.05. HCl is limiting. From stoichiometry: 6 mol HCl → 3 mol H₂, so 0.30 mol HCl → 0.15 mol H₂. Volume at STP = 0.15 × 22.4 = 3.36 L.
Why A is wrong: A is wrong because 4.48 L = 0.20 × 22.4, which assumes Al is limiting and uses its full moles to produce H₂ without checking HCl (trap: skipping limiting reagent check and using the first reactant directly).
Why B is wrong: B is wrong because 2.24 L = 0.10 × 22.4, which likely comes from using the Al ratio (0.10) directly as moles of H₂ without applying the 2:3 coefficient relationship between Al and H₂ (trap: confusing the limiting-reagent ratio with product moles).
Why C is wrong: C is wrong because 6.72 L = 0.30 × 22.4, which treats all HCl moles as producing H₂ in a 1:1 ratio instead of the correct 6:3 stoichiometry (trap: using wrong mole ratio between HCl and H₂).
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)
Show answer and why every option is right or wrong
Answer: C. Effective mass = 10.0 × 0.75 = 7.50 g. Moles MgCO₃ = 7.50/84 = 0.08929 mol. From 1:1 stoichiometry, moles CO₂ = 0.08929 mol. Mass CO₂ = 0.08929 × 44 = 3.93 g (traps: percentage purity + stoichiometric mole-mass conversion).
Why A is wrong: A is wrong because 5.24 g comes from 10.0/84 × 44 = 5.24, which uses the total impure sample mass without applying the 75% purity correction (trap: forgetting percentage purity step).
Why B is wrong: B is wrong because 4.40 g likely comes from 0.10 × 44 = 4.40, using 10.0/100 = 0.10 mol which incorrectly divides by 100 instead of the molar mass 84 (trap: using sample mass ÷ 100 instead of ÷ molar mass).
Why D is wrong: D is wrong because 7.50 g is the effective pure mass of MgCO₃ (10 × 0.75), not the mass of CO₂ produced — you stopped after the purity step without completing the mole-to-mass conversion (trap: reporting intermediate value as final answer).
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Stoichiometry: quick recall before you leave
How do you solve a Stoichiometry question? A 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)
---
What to remember before solving Stoichiometry questions
Limiting reagent
1.10.1 Limiting Reagent Many a time, reactions are carried out with the amounts of reactants that are different than the amounts as required by a balanced chemical reaction. In such situations, one reactant is in more amount than the amount required by balanced chemical reaction. The reactant which is present in the least amount gets consumed after sometime and after that further reaction does not take place whatever be the amount of the other reactant. Hence, the reactant, which gets consumed first, limits the amount of product formed and is, therefore, called the limiting reagent. In performing stoichiometric calculations, this aspect is also to be kept in mind.
-- NCERT Class 11 Chemistry, Ch. 1, p. 20Concentration in solutions
2. Mole Fraction It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB, respectively, then the mole fractions of A and B are given as: Mole fraction of A = No. of moles of A / No. of moles of solutions = nA / (nA + nB) Mole fraction of B = No. of moles of B / No. of moles of solutions = nB / (nA + nB) 3. Molarity It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution. Thus, Molarity (M) = No. of moles of solute / Volume of solution in litres … Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent. 4. Molality It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m. Thus, Molality (m) = No. of moles of solute / Mass of solvent in kg … Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature.
-- NCERT Class 11 Chemistry, Ch. 1, p. 23Which Stoichiometry formulas do you need for NEET?
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 273.15 K and 1 bar.
| 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
- 22.7 L mol^-1 only at 273.15 K and 1 bar (22.4 L mol^-1 if the question uses 1 atm); otherwise PV = nRT
- Pure substance
Where do students lose marks on Stoichiometry?
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 a fixed molar volume (22.7 L/mol or 22.4 L/mol) at conditions it does not apply to, or mixes the two conventions.
When it triggers
Question gives gas at non-STP T or P.
How to avoid
One mole of an ideal gas occupies 22.7 L at 273.15 K and 1 bar (100 kPa), the value NCERT's reprint uses; 22.4 L is the older figure for 273.15 K and 1 atm (101.325 kPa). Use whichever matches the conditions the question states; for any other T or P, 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
Effective mass = stated mass × (purity/100). Then convert to moles using molar mass.
More in Some Basic Concepts of Chemistry: 1 exam trap or mistake from its other lessons.
Stoichiometry questions from past NEET papers
4 questions from NEET 2022, 2023, 2024, 2026. Answers verified against NTA official keys.
All 12 past-paper questions from Some Basic Concepts of Chemistry →
How does NEET ask about Stoichiometry?
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.
Common distractors
swapped classes
Tempts surface-level recall.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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