Stoichiometry

Complete guide to stoichiometric calculations, limiting reagent, excess reagent, and percentage yield for JEE Main and Advanced preparation

13 min read

Stoichiometry

Real-Life Hook: The Perfect Recipe for Pancakes

Imagine making pancakes: 1 cup flour + 2 eggs + 1 cup milk = 4 pancakes. What if you have 3 cups flour, 10 eggs, and 2 cups milk? You can only make 8 pancakes because milk runs out first! This is exactly what chemists call a “limiting reagent” problem. In chemical reactions, just like cooking, you need ingredients in the right proportions - and one ingredient always runs out first, limiting how much product you can make.

Introduction

Stoichiometry (from Greek: “stoicheion” = element, “metron” = measure) is the calculation of reactants and products in chemical reactions. It’s the mathematical heart of chemistry, essential for:

  • Predicting how much product forms
  • Determining which reactant limits the reaction
  • Calculating percentage yield and purity
  • Industrial chemical process optimization

For JEE, stoichiometry problems are among the most frequently tested topics, appearing in multiple choice, numerical, and assertion-reason formats.

Balanced Chemical Equations

Law of Conservation of Mass

In a chemical reaction, the total mass of reactants equals the total mass of products.

$$\boxed{\text{Mass of reactants} = \text{Mass of products}}$$

This requires balanced equations where:

  • Number of atoms of each element is equal on both sides
  • Total charge is conserved (in ionic equations)

Balancing Equations - Strategy

Step-by-step approach:

  1. Write skeleton equation with correct formulas
  2. Count atoms of each element on both sides
  3. Balance metals first, then non-metals, then H and O
  4. Use whole number coefficients (multiply if needed to clear fractions)
  5. Verify atom count and charges

Example: Balancing

Balance: Fe + O₂ → Fe₂O₃

Step 1: Skeleton Fe + O₂ → Fe₂O₃

Step 2: Count

  • Left: Fe = 1, O = 2
  • Right: Fe = 2, O = 3

Step 3: Balance Fe first 2Fe + O₂ → Fe₂O₃

Step 4: Balance O 2Fe + 3/2 O₂ → Fe₂O₃

Step 5: Clear fraction (multiply all by 2)

$$\boxed{4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3}$$

Stoichiometric Coefficients

Mole Ratio

The coefficients in a balanced equation represent the mole ratio of reactants and products.

For:

$$N_2 + 3H_2 \rightarrow 2NH_3$$

Mole ratios:

  • 1 mole N₂ : 3 moles H₂ : 2 moles NH₃
  • N₂ : H₂ : NH₃ = 1 : 3 : 2

What Coefficients Tell Us

InterpretationN₂ + 3H₂ → 2NH₃
Molecules1 molecule N₂ + 3 molecules H₂ → 2 molecules NH₃
Moles1 mol N₂ + 3 mol H₂ → 2 mol NH₃
Mass28 g N₂ + 6 g H₂ → 34 g NH₃
Volume (gas, STP)22.4 L N₂ + 67.2 L H₂ → 44.8 L NH₃

Memory Trick: “Coefficient = Mole Recipe”

Think of coefficients as a mole recipe:

  • 1 mol N₂ + 3 mol H₂ → 2 mol NH₃
  • Just like: 1 egg + 3 cups flour → 2 cakes

Basic Stoichiometric Calculations

General Approach

Given amount → Moles → Mole ratio → Moles → Desired quantity
   (÷M)         of A     (from eq)    of B      (×M or ×Nₐ)

Formula Framework

$$\boxed{\frac{n_A}{a} = \frac{n_B}{b}}$$

Where:

  • nₐ, nᵦ = moles of substances A and B
  • a, b = stoichiometric coefficients

Interactive Demo: Visualize Mole Concept

Explore mole ratios and stoichiometric calculations interactively.

Solved Example 1: Mass-to-Mass

Calculate mass of NH₃ produced from 14 g of N₂. Reaction: N₂ + 3H₂ → 2NH₃

Solution:

Step 1: Find moles of N₂

$$n_{N_2} = \frac{14}{28} = 0.5 \text{ mol}$$

Step 2: Use mole ratio

$$\frac{n_{N_2}}{1} = \frac{n_{NH_3}}{2}$$ $$n_{NH_3} = 2 \times 0.5 = 1 \text{ mol}$$

Step 3: Convert to mass

$$m_{NH_3} = 1 \times 17 = 17 \text{ g}$$ $$\boxed{m_{NH_3} = 17 \text{ g}}$$

Solved Example 2: Volume-to-Mass (Gas)

What mass of O₂ is required to react with 11.2 L of H₂ at STP? Reaction: 2H₂ + O₂ → 2H₂O

Solution:

Step 1: Find moles of H₂

$$n_{H_2} = \frac{11.2}{22.4} = 0.5 \text{ mol}$$

Step 2: Use mole ratio

$$\frac{n_{H_2}}{2} = \frac{n_{O_2}}{1}$$ $$n_{O_2} = \frac{0.5}{2} = 0.25 \text{ mol}$$

Step 3: Convert to mass

$$m_{O_2} = 0.25 \times 32 = 8 \text{ g}$$ $$\boxed{m_{O_2} = 8 \text{ g}}$$

Limiting Reagent

Definition

The limiting reagent (or limiting reactant) is the reactant that is completely consumed first, limiting the amount of product formed.

The excess reagent is the reactant that remains after the reaction is complete.

Analogy: Car Manufacturing

To make 1 car: 1 body + 4 wheels

If you have:

  • 10 bodies
  • 30 wheels

Maximum cars = 30/4 = 7.5 → 7 cars (can’t make half!)

  • Wheels are limiting (need 40 for 10 cars, have only 30)
  • Bodies are in excess (3 bodies left over)

How to Find Limiting Reagent

Method 1: Mole Ratio Method

  1. Calculate moles of each reactant
  2. Divide moles by stoichiometric coefficient
  3. Smallest value → limiting reagent

Method 2: Product Method

  1. Calculate product from each reactant (assuming others are excess)
  2. Reactant giving least product → limiting reagent

Solved Example 3: Limiting Reagent

10 g N₂ reacts with 10 g H₂. Find: a) Limiting reagent b) Mass of NH₃ formed c) Excess reagent remaining

Reaction: N₂ + 3H₂ → 2NH₃

Solution:

Part a) Find limiting reagent

Moles of N₂ = 10/28 = 0.357 mol Moles of H₂ = 10/2 = 5 mol

Using mole ratio method:

$$\frac{n_{N_2}}{1} = \frac{0.357}{1} = 0.357$$ $$\frac{n_{H_2}}{3} = \frac{5}{3} = 1.667$$

Smallest value → N₂ is limiting reagent

Part b) Mass of NH₃

From limiting reagent N₂:

$$\frac{0.357}{1} = \frac{n_{NH_3}}{2}$$ $$n_{NH_3} = 0.714 \text{ mol}$$ $$m_{NH_3} = 0.714 \times 17 = 12.14 \text{ g}$$

Part c) Excess reagent

H₂ required = 3 × 0.357 = 1.071 mol = 2.142 g H₂ in excess = 10 - 2.142 = 7.858 g

$$\boxed{\text{a) N}_2 \text{ limiting; b) 12.14 g NH}_3; \text{c) 7.86 g H}_2 \text{ excess}}$$

Percentage Yield

Theoretical vs Actual Yield

  • Theoretical yield: Maximum product possible (calculated from stoichiometry)
  • Actual yield: Product actually obtained (from experiment)
  • Percentage yield: Efficiency of reaction
$$\boxed{\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%}$$

Why Actual < Theoretical?

  1. Incomplete reactions: Equilibrium, reversible reactions
  2. Side reactions: Formation of byproducts
  3. Loss during separation: Filtration, crystallization losses
  4. Impure reactants: Less than 100% purity
  5. Experimental errors: Spillage, measurement errors

Solved Example 4: Percentage Yield

20 g of CaCO₃ is heated. 8.8 g of CO₂ is collected. Calculate percentage yield. Reaction: CaCO₃ → CaO + CO₂

Solution:

Step 1: Theoretical yield

Moles of CaCO₃ = 20/100 = 0.2 mol From equation: 1 mol CaCO₃ → 1 mol CO₂ Moles of CO₂ = 0.2 mol Theoretical mass of CO₂ = 0.2 × 44 = 8.8 g

Wait, this matches! Let me recalculate with actual < theoretical:

Actually, if 8.8 g is collected and theoretical is also 8.8 g:

$$\text{Percentage yield} = \frac{8.8}{8.8} \times 100 = 100\%$$

For a more realistic problem, let’s say 7 g was collected:

$$\text{Percentage yield} = \frac{7}{8.8} \times 100 = 79.5\%$$

Consecutive Reactions

When reactions occur in sequence: A → B → C

Calculate step by step, using product of first reaction as reactant for second.

Solved Example 5: Consecutive Reactions

$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$ $$\text{CO}_2 + \text{C} \rightarrow 2\text{CO}$$

If 12 g of C is burnt, then CO₂ reacts with 6 g more C, find total mass of CO formed.

Solution:

Step 1: First reaction

$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$

Moles of C = 12/12 = 1 mol Moles of CO₂ = 1 mol Mass of CO₂ = 44 g

Step 2: Second reaction

$$\text{CO}_2 + \text{C} \rightarrow 2\text{CO}$$

Moles of CO₂ = 44/44 = 1 mol (from step 1) Moles of C = 6/12 = 0.5 mol (given)

Check limiting reagent:

  • From CO₂: 1 mol → 2 mol CO
  • From C: 0.5 mol → 1 mol CO

C is limiting! Moles of CO = 1 mol (from 0.5 mol C)

Mass of CO = 1 × 28 = 28 g

$$\boxed{m_{CO} = 28 \text{ g}}$$

Common Calculation Mistakes

Mistake 1: Using Mass Ratio Instead of Mole Ratio

❌ Wrong: For N₂ + 3H₂ → 2NH₃, if 28 g N₂ reacts with 6 g H₂ Using mass: Product = 28 + 6 = 34 g (correct by luck!)

✓ Right: Always convert to moles first, use mole ratio, then convert back

Mistake 2: Not Finding Limiting Reagent

❌ Wrong: Calculating product from each reactant separately without checking which limits

✓ Right: Always identify limiting reagent first in excess reactant problems

Mistake 3: Using Unbalanced Equations

❌ Wrong: Fe + O₂ → Fe₂O₃ (unbalanced!)

✓ Right: 4Fe + 3O₂ → 2Fe₂O₃ (balanced)

Mistake 4: Wrong Molar Masses

❌ Wrong: M(O₂) = 16 g/mol (atomic mass)

✓ Right: M(O₂) = 32 g/mol (molecular mass)

Mistake 5: Percentage Yield > 100%

❌ Wrong: If calculated yield > 100%, accepting it

✓ Right: Check calculations - yield can’t exceed 100% (impure product possible)

Memory Tricks for Stoichiometry

“MOLEMOLE” Method

  1. Mass to moles (given substance) - divide by M
  2. Observe the balanced equation
  3. Look for mole ratio
  4. Equate the ratios: nₐ/a = nᵦ/b
  5. Moles of unknown calculated
  6. Obtain desired quantity
  7. Label with units
  8. Evaluate if reasonable

Limiting Reagent Trick: “Divide and Conquer”

Divide moles by coefficient Conquer by choosing smallest

SYRUP for Yield Problems

  • Start with theoretical calculation
  • Yield actual from experiment
  • Ratio: actual/theoretical
  • Units: percentage
  • Practical is always less

Practice Problems

Level 1: Basic (JEE Main)

Q1. Balance the equation: Al + O₂ → Al₂O₃

Answer4Al + 3O₂ → 2Al₂O₃

Check: Al: 4 = 4 ✓, O: 6 = 6 ✓

Q2. How many moles of O₂ are needed to react completely with 2 moles of C? Reaction: C + O₂ → CO₂

SolutionFrom balanced equation: 1 mol C : 1 mol O₂

For 2 mol C → 2 mol O₂ needed

Answer: 2 moles O₂

Q3. Calculate mass of H₂O formed from 4 g H₂. Reaction: 2H₂ + O₂ → 2H₂O

SolutionMoles of H₂ = 4/2 = 2 mol

From equation: 2 mol H₂ → 2 mol H₂O So 2 mol H₂ → 2 mol H₂O

Mass of H₂O = 2 × 18 = 36 g

Answer: 36 g

Q4. What volume of CO₂ (at STP) is produced from 50 g CaCO₃? Reaction: CaCO₃ → CaO + CO₂

SolutionMoles of CaCO₃ = 50/100 = 0.5 mol

From equation: 1 mol CaCO₃ → 1 mol CO₂ So 0.5 mol CaCO₃ → 0.5 mol CO₂

Volume of CO₂ = 0.5 × 22.4 = 11.2 L

Answer: 11.2 L at STP

Level 2: Intermediate (JEE Main/Advanced)

Q5. 14 g of N₂ and 6 g of H₂ are mixed and reacted. Find: a) Limiting reagent b) Mass of NH₃ formed Reaction: N₂ + 3H₂ → 2NH₃

SolutionMoles of N₂ = 14/28 = 0.5 mol Moles of H₂ = 6/2 = 3 mol

Checking limiting reagent: n(N₂)/1 = 0.5/1 = 0.5 n(H₂)/3 = 3/3 = 1

N₂ is limiting (smaller value)

From N₂: 0.5 mol N₂ → 1 mol NH₃ Mass of NH₃ = 1 × 17 = 17 g

Answer: a) N₂ limiting, b) 17 g NH₃

Q6. 10 g of impure CaCO₃ produces 2.24 L of CO₂ at STP. Calculate % purity of CaCO₃. Reaction: CaCO₃ → CaO + CO₂

SolutionMoles of CO₂ = 2.24/22.4 = 0.1 mol

From equation: 1 mol CO₂ comes from 1 mol CaCO₃ Moles of pure CaCO₃ = 0.1 mol Mass of pure CaCO₃ = 0.1 × 100 = 10 g

% purity = (10/10) × 100 = 100%

Answer: 100% pure

(Note: If sample was impure, the given mass would be greater than pure mass calculated)

Q7. A mixture of FeO and Fe₂O₃ when heated with carbon monoxide, produces 11.2 g of Fe. If the mixture contained 0.15 mol of FeO, find moles of Fe₂O₃.

Reactions: FeO + CO → Fe + CO₂ Fe₂O₃ + 3CO → 2Fe + 3CO₂

SolutionMoles of Fe produced = 11.2/56 = 0.2 mol

From FeO: 0.15 mol FeO → 0.15 mol Fe

Remaining Fe must come from Fe₂O₃: Moles of Fe from Fe₂O₃ = 0.2 - 0.15 = 0.05 mol

From equation: 1 mol Fe₂O₃ → 2 mol Fe Moles of Fe₂O₃ = 0.05/2 = 0.025 mol

Answer: 0.025 mol Fe₂O₃

Q8. Theoretical yield of a reaction is 25 g. If 20 g is obtained, what is the percentage yield?

Solution% yield = (20/25) × 100 = 80%

Answer: 80%

Level 3: Advanced (JEE Advanced - Tricky)

Q9. (JEE Advanced Pattern) A gaseous mixture of H₂ and CO in molar ratio a:b reacts with O₂ to form H₂O and CO₂. If the ratio of O₂ consumed to H₂O formed is 5:4, find a:b.

Reactions: 2H₂ + O₂ → 2H₂O 2CO + O₂ → 2CO₂

SolutionLet moles: H₂ = a, CO = b

From H₂ reaction: a mol H₂ consumes a/2 mol O₂, produces a mol H₂O From CO reaction: b mol CO consumes b/2 mol O₂, produces b mol CO₂

Total O₂ consumed = a/2 + b/2 = (a+b)/2 Total H₂O formed = a

Given: O₂ : H₂O = 5 : 4

$$\frac{(a+b)/2}{a} = \frac{5}{4}$$ $$\frac{a+b}{2a} = \frac{5}{4}$$ $$4(a+b) = 10a$$ $$4a + 4b = 10a$$ $$4b = 6a$$ $$\frac{a}{b} = \frac{4}{6} = \frac{2}{3}$$

Answer: a:b = 2:3

Q10. (JEE Advanced 2017 Pattern) When 10 g of a metal M reacts with excess HCl, 11.2 L of H₂ (at STP) is evolved. The equivalent weight of metal is: a) 10 b) 20 c) 30 d) 40

SolutionMoles of H₂ = 11.2/22.4 = 0.5 mol

Reaction: M + nHCl → MCl_n + (n/2)H₂

For 1 equivalent of metal, 1/2 equivalent of H₂ (or 11.2 L) is produced

Equivalents of H₂ = 0.5 × 2 = 1 equivalent (since H₂ has 2 H, each H⁺ is 1 equivalent)

Actually, simpler approach: 1 mol H₂ requires 2 equivalents of metal 0.5 mol H₂ requires 1 equivalent of metal

Equivalent weight = mass/equivalents = 10/1 = 10

Answer: a) 10

Q11. (Tricky) In the reaction C₃H₈ + O₂ → CO₂ + H₂O, if 22 g of C₃H₈ is burnt with 80 g of O₂: a) Which is limiting reagent? b) How much excess reagent remains? c) Mass of CO₂ formed?

SolutionFirst, balance the equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Moles of C₃H₈ = 22/44 = 0.5 mol Moles of O₂ = 80/32 = 2.5 mol

Check limiting: n(C₃H₈)/1 = 0.5/1 = 0.5 n(O₂)/5 = 2.5/5 = 0.5

Both exactly react! No limiting reagent, both are completely consumed.

a) No limiting reagent (both completely react)

b) No excess (both completely consumed)

c) Mass of CO₂: From 0.5 mol C₃H₈ → 1.5 mol CO₂ Mass = 1.5 × 44 = 66 g

Answer: a) None, b) 0 g, c) 66 g CO₂

Q12. (Conceptual) Two elements X and Y form compounds XY and X₂Y₃. If 0.1 mol of XY weighs 10 g and 0.05 mol of X₂Y₃ weighs 12.5 g, find atomic masses of X and Y.

SolutionM(XY) = 10/0.1 = 100 g/mol M(X₂Y₃) = 12.5/0.05 = 250 g/mol

Let atomic mass of X = x, Y = y

From XY: x + y = 100 … (1) From X₂Y₃: 2x + 3y = 250 … (2)

From (1): x = 100 - y Substitute in (2): 2(100 - y) + 3y = 250 200 - 2y + 3y = 250 y = 50

From (1): x = 100 - 50 = 50

Answer: X = 50 amu, Y = 50 amu

For JEE Advanced Students

  • Gravimetric analysis
  • Volumetric analysis (titrations)
  • Eudiometry
  • Combustion analysis for formula determination
  • Industrial applications (Haber process, Contact process)

Quick Revision Points

Stoichiometry Steps

  1. Write balanced equation
  2. Convert given to moles
  3. Use mole ratio
  4. Convert to desired units
  5. Check reasonableness

Limiting Reagent

  • Divide moles by coefficient
  • Smallest value = limiting
  • Calculate products from limiting reagent only
  • Excess = initial - consumed

Important Formulas

$$\boxed{\frac{n_A}{a} = \frac{n_B}{b}}$$

(mole ratio)

$$\boxed{\text{% Yield} = \frac{\text{Actual}}{\text{Theoretical}} \times 100}$$ $$\boxed{\text{Mass conserved: } \sum m_{\text{reactants}} = \sum m_{\text{products}}}$$

Tips for JEE

  1. Always balance first - No calculations without balanced equation
  2. Show all steps - Partial credit in numerical problems
  3. Unit consistency - Keep track of g, mol, L throughout
  4. Limiting reagent - Default assumption: find it first
  5. Reasonableness check - Does answer make sense?
  6. Gas problems - Remember 22.4 L/mol at STP
  7. Percentage yield - Never exceeds 100% in reality
  8. Practice variety - Mass-mass, mass-volume, volume-volume problems

Last updated: May 2025 Part of JEE Chemistry - Basic Concepts series