ChemistryCBSE Class 11 · Some Basic Concepts of Chemistry

Stoichiometry, Limiting Reagent and Solution Concentration Class 11

Updated 17 October 2026 · Fact-checked

Stoichiometry uses the coefficients of a balanced equation as mole ratios. The reagent that gives the least product is the limiting reagent; percentage yield = actual/theoretical × 100. Concentrations: molarity = moles of solute per litre of solution, molality = moles per kg of solvent, mole fraction = moles of a component per total moles, and dilution obeys M₁V₁ = M₂V₂.

Divide each reactant's moles by its coefficient; the smallest value (O₂ = 2) is the limiting reagent and decides how much product forms.

Illustration drawn for QuestionBuddy.

Understand Stoichiometry, Limiting Reagent and Solution Concentration

Mole ratios. A balanced equation such as $CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O$ gives mole ratios (1 : 2 : 1 : 2), not mass ratios. Method: mass → moles → mole ratio → moles of the wanted species → mass (or volume of solution).

Limiting reagent. Divide the moles of each reactant by its coefficient; the reactant with the smallest result is limiting and fixes the amount of product. The other reactant is in excess; the unreacted amount is (moles supplied) − (moles used).

Yield and purity. $% yield = \frac{actual yield}{theoretical yield} \times 100$ . For an impure sample, use only the mass of the pure component (mass × % purity).

Concentration terms.

Mass percent $= \frac{m _{so l u t e}}{m _{so l u t i o n}} \times 100$ .
Mole fraction $x_{A} = \frac{n _{A}}{n _{A} + n _{B}}$ ; $\sum x = 1$ .
Molarity $M = \frac{n _{so l u t e}}{V _{so l u t i o n} ( L )}$ (changes with temperature).
Molality $m = \frac{n _{so l u t e}}{m _{so l v e n t} ( kg )}$ (independent of temperature).
From density $d$ (g/mL) and mass %: $M = \frac{10 d ( % )}{M _{so l u t e}}$ .
Dilution: $M_{1} V_{1} = M_{2} V_{2}$ . For neutralisation, moles of H⁺ equal moles of OH⁻.

Back titration. A known excess of reagent is added to a sample, and the leftover is titrated; moles used = moles added − moles left.

Mixing solutions. Add moles of each ion, divide by the total volume (assuming volumes add).

Stoichiometry, Limiting Reagent and Solution Concentration: formulas and key facts

Limiting reagent: $\frac{n _{r e a c t an t}}{coefficient} = minimum$; Smallest ratio limits.
Percentage yield: $\frac{actual}{theoretical} \times 100$; Efficiency.
Molarity: $M = \frac{n}{V ( L )}$; Per litre of solution.
Molality: $m = \frac{n}{m _{so l v e n t} ( kg )}$; Independent of T.
Molarity from mass %: $M = \frac{10 d ( % )}{M _{so l u t e}}$; d in g/mL.
Dilution: $M_{1} V_{1} = M_{2} V_{2}$; Moles unchanged.

How to solve Stoichiometry, Limiting Reagent and Solution Concentration questions

Convert to moles, use the coefficients, find the limiting reagent, then convert back to the unit asked.

1Balance the equation and check atom counts.
2Convert each given mass or volume of solution to moles.
3Divide each by its coefficient to identify the limiting reagent.
4Use the limiting reagent for the moles of product; subtract to find the excess left.
5Convert to mass, and apply yield or purity if given.
6For concentration, keep track of moles of solute, volume of solution and mass of solvent separately.

Stoichiometry, Limiting Reagent and Solution Concentration in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Diluting concentrated acid

Suppose a lab technician has 6.0 M HCl and needs 250 mL of 0.30 M HCl.

What volume of the concentrated acid is needed?

Try it first, then show the solutionSolution

M₁V₁ = M₂V₂: V₁ = 0.30 × 250/6.0 = 12.5 mL.

Answer: 12.5 mL, made up to 250 mL with water.

Memory hook: The moles of HCl stay the same.

Baking soda and vinegar

Suppose a student mixes 4.2 g of NaHCO₃ with excess acetic acid: NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂.

Find the mass of CO₂ released (NaHCO₃ = 84 g/mol, CO₂ = 44 g/mol).

Try it first, then show the solutionSolution

n(NaHCO₃) = 4.2/84 = 0.050 mol; mole ratio 1 : 1.
m(CO₂) = 0.050 × 44 = 2.2 g.

Answer: 2.2 g of CO₂.

Memory hook: The coefficients are all 1.

Tricks to remember Stoichiometry, Limiting Reagent and Solution Concentration

Short memory aids. Say each one aloud once, then use it on a question the same day.

Divide by coefficient
The smallest moles/coefficient identifies the limiting reagent.
Use it when: Limiting reagent.
Molality uses solvent
Kilograms of solvent, not of solution.
Use it when: Concentration terms.
M = 10 d % / M
Quick molarity from density and percentage.
Use it when: Concentrated acids.
Moles are conserved on dilution
M₁V₁ = M₂V₂.
Use it when: Dilution.

Stoichiometry, Limiting Reagent and Solution Concentration: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Mass of product · 2 marks · Very short answerBoard level

Find the mass of carbon dioxide produced when 8.0 g of methane burns completely in excess oxygen: CH₄ + 2O₂ → CO₂ + 2H₂O (C = 12, H = 1, O = 16).

Try it first, then show the solutionSolution

n(CH₄) = 8.0/16 = 0.50 mol; ratio CH₄ : CO₂ = 1 : 1. (1)
m(CO₂) = 0.50 × 44 = 22 g. (1)

Answer: 22 g.

Example 2: Concentrated hydrochloric acid · 3 marks · Short answerHarder board level

A concentrated hydrochloric acid is 36.5% HCl by mass and has density 1.18 g/mL (HCl = 36.5 g/mol). Find its molarity and molality, and the volume needed to prepare 500 mL of 0.10 M HCl.

Try it first, then show the solutionSolution

In 1 L: mass of solution = 1180 g, HCl = 430.7 g = 11.8 mol, so M = 11.8 mol/L. (1)
Solvent = 1180 − 430.7 = 749.3 g, so m = 11.8/0.7493 = 15.7 mol/kg. (1)
V = 0.10 × 500/11.8 = 4.2 mL. (1)

Answer: 11.8 M; 15.7 m; 4.2 mL.

Example 3: Limiting reagent · 3 marks · Short answerHarder board level

5.4 g of aluminium reacts with 12.0 g of chlorine: 2Al + 3Cl₂ → 2AlCl₃ (Al = 27, Cl = 35.5). Find the limiting reagent, the mass of AlCl₃ formed and the mass of the excess reagent left.

Try it first, then show the solutionSolution

n(Al) = 5.4/27 = 0.200 mol; n(Cl₂) = 12.0/71 = 0.169 mol. Al/2 = 0.100, Cl₂/3 = 0.0563, so Cl₂ is limiting. (1)
n(AlCl₃) = (2/3) × 0.169 = 0.113 mol; M = 133.5, mass = 15.0 g. (1)
Al used = 0.113 mol; left = 0.200 − 0.113 = 0.0873 mol = 2.36 g. (1)

Answer: Cl₂ limits; 15.0 g of AlCl₃; 2.36 g of Al left.

Example 4: Yield from impure limestone · 4 marks · Long answerHarder board level

50 g of limestone containing 90% CaCO₃ is heated: CaCO₃ → CaO + CO₂ (Ca = 40, C = 12, O = 16). The CaO obtained weighs 21.0 g. Find the theoretical yield of CaO, the percentage yield and the mass of CO₂ evolved theoretically.

Try it first, then show the solutionSolution

Pure CaCO₃ = 0.90 × 50 = 45 g = 0.450 mol. (1)
Theoretical CaO = 0.450 × 56 = 25.2 g. (1)
Percentage yield = 21.0/25.2 × 100 = 83.3%. (1)
Theoretical CO₂ = 0.450 × 44 = 19.8 g. (1)

Answer: 25.2 g; 83.3%; 19.8 g.

Example 5: Back titration · 4 marks · Long answerHOTS / competency

1.00 g of impure calcium carbonate is treated with 50.0 mL of 0.50 M HCl. The excess acid needs 24.0 mL of 0.25 M NaOH. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Find the percentage purity of the sample and the mass of CO₂ evolved.

Try it first, then show the solutionSolution

HCl added = 0.050 × 0.50 = 25.0 mmol; excess = 24.0 × 0.25 = 6.0 mmol (1 : 1 with NaOH). (1)
HCl used = 19.0 mmol. (1)
n(CaCO₃) = 19.0/2 = 9.5 mmol = 0.95 g, purity 95%. (1)
n(CO₂) = 9.5 mmol, mass = 0.418 g. (1)

Answer: 95% pure; 0.418 g of CO₂.

Where students lose marks in Stoichiometry, Limiting Reagent and Solution Concentration

The slips examiners see most often, and what to write instead.

Mistake: Writing an unbalanced equation.
Skipping the balance check.
Do this: Balance first; coefficients are mole ratios.
Usually costs: The whole numerical.
Mistake: Treating the larger mass as the limiting reagent.
Comparing masses.
Do this: Compare moles ÷ coefficient.
Usually costs: The answer mark.
Mistake: Using the mass of solution in the molality.
Wrong denominator.
Do this: Molality uses the mass of the solvent in kg.
Usually costs: 1 mark.
Mistake: Using 100% purity for an impure sample.
Ignoring the purity.
Do this: Multiply the mass by the fractional purity.
Usually costs: Method mark.

Exam tips

Write the balanced equation first and show its atom count.
Tabulate moles, coefficient and ratio for the limiting reagent.
State clearly which volume (solution or solvent) is used.
Convert mL to L and g to kg before substituting.
Report percent yield to three significant figures.

Practice questions on Stoichiometry, Limiting Reagent and Solution Concentration

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
4.0 g of sodium hydroxide is dissolved in water to make 500 mL of solution (NaOH = 40 g/mol). The molarity is
(a) 0.1 M (b) 0.2 M (c) 0.4 M (d) 2.0 M
Hint
n = 0.10 mol in 0.5 L.
Answer
(b) 0.2 M.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): One litre of 1 M glucose solution and one litre of 1 M urea solution contain the same number of solute molecules.
Reason (R): Equal volumes of solutions of equal molarity contain equal moles of solute.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Moles = M × V.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
100 mL of 0.50 M sodium chloride solution is diluted to 250 mL. Find the new molarity.
Hint
M₁V₁ = M₂V₂.
Answer
M₂ = 0.50 × 100/250 = 0.20 M.
Q4 · 3 marks · Short answerHarder board level
A solution is made by mixing 36 g of water with 46 g of ethanol, C₂H₅OH (H₂O = 18, C₂H₅OH = 46 g/mol). Find the mole fraction of each, the mass percent of ethanol and the molality of ethanol in water.
Hint
n(water) = 2.0; n(ethanol) = 1.0.
Answer
x(ethanol) = 1/3 = 0.333, x(water) = 0.667; mass % ethanol = 46/82 × 100 = 56.1%; molality = 1.0/0.036 = 27.8 mol/kg.
Q5 · 4 marks · Long answerHOTS / competency
100 mL of 0.20 M BaCl₂ is mixed with 100 mL of 0.30 M Na₂SO₄: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl (BaSO₄ = 233 g/mol). Find the mass of precipitate and the concentrations of Na⁺, Cl⁻ and the leftover SO₄²⁻ in the filtrate (volumes add).
Hint
Ba²⁺ is limiting.
Answer
n(Ba²⁺) = 0.020 mol, n(SO₄²⁻) = 0.030 mol; BaSO₄ = 0.020 mol = 4.66 g. Leftover SO₄²⁻ = 0.010 mol in 0.200 L = 0.050 M; Na⁺ = 0.060 mol/0.200 L = 0.30 M; Cl⁻ = 0.040/0.200 = 0.20 M.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Limiting reagent test?Show answer
Smallest moles ÷ coefficient.
2.Percentage yield?Show answer
Actual/theoretical × 100.
3.Molality uses?Show answer
Mass of solvent in kg.
4.Dilution law?Show answer
M₁V₁ = M₂V₂.
5.Mole fractions add to?Show answer
1.
6.Molarity from d and %?Show answer
10 d % / M.

Stoichiometry, Limiting Reagent and Solution Concentration: frequently asked questions

Why does molarity change with temperature but molality does not?

Volume changes with temperature, while mass does not.

What is a limiting reagent?

The reactant that is completely consumed first and therefore limits the amount of product.

Can percentage yield exceed 100%?

Not for a pure product; a value above 100% usually means impurities or a mistake.

What is a back titration?

Adding excess reagent and titrating the leftover to find how much reacted.

Why is dilution calculated with M₁V₁ = M₂V₂?

Because the moles of solute stay the same, only the volume changes.