All important stoichiometry notes on a few pages pdf

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Complete Stoichiometry Guide – All Concepts on Few Pages

All important stoichiometry notes on a few pages

All essential concepts condensed into a powerful reference guide

Mole Concept Balancing Equations Limiting Reactants Solution Stoichiometry NEET/JEE Focus

Why Stoichiometry Matters

Stoichiometry is the quantitative study of reactants and products in chemical reactions. It’s the foundation for:

  • Predicting product amounts in chemical reactions
  • Determining limiting reactants
  • Calculating reaction yields
  • Understanding solution concentrations
  • Industrial chemical process design

NEET/JEE Importance

Stoichiometry accounts for ~15% of physical chemistry questions in competitive exams. Master these concepts to gain a significant advantage.

Core Stoichiometry Concepts

1. The Mole Concept

One mole = 6.022 × 10²³ particles (Avogadro’s number)

This applies to atoms, molecules, ions, or any chemical entity.

Key Formulas:

\[ n = \frac{N}{N_A} \]

Where:
n = amount of substance (mol)
N = number of particles
NA = Avogadro’s constant

\[ n = \frac{m}{M} \]

Where:
m = mass (g)
M = molar mass (g/mol)

Example: How many atoms are in 2.5 moles of iron?

Solution:

Number of atoms = moles × Avogadro’s number
= 2.5 mol × 6.022 × 10²³ atoms/mol
= 1.5055 × 10²⁴ atoms

2. Balancing Chemical Equations

5-Step Balancing Method:

  1. Write the unbalanced equation
  2. Count atoms of each element on both sides
  3. Add coefficients to balance atoms (start with complex molecules)
  4. Balance hydrogen and oxygen last
  5. Verify all atoms are balanced
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

Unbalanced combustion of propane

\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]

Balanced equation

Exam Tip:

For redox reactions, use the half-reaction method. Always double-check your balanced equations by counting all atoms!

3. Stoichiometric Calculations

The stoichiometric roadmap:

Mass A → Moles A → Moles B → Mass B

General Approach:

Given → Moles

Convert given quantity to moles using appropriate conversion factor (molar mass for mass, molarity for solutions, etc.)

Mole Ratio

Use coefficients from balanced equation to convert between moles of different substances

Example: What mass of CO2 is produced from burning 50g of propane (C3H8)?

Solution Steps:

  1. Write balanced equation: C3H8 + 5O2 → 3CO2 + 4H2O
  2. Convert 50g C3H8 to moles: n = m/M = 50g / 44.1g/mol = 1.134 mol
  3. Mole ratio (from equation): 1 mol C3H8 : 3 mol CO2
  4. Moles CO2 = 1.134 mol × 3 = 3.402 mol
  5. Convert to mass: m = n × M = 3.402 mol × 44.0g/mol = 149.7g CO2

4. Limiting Reactant Theory

Limiting Reactant

The reactant that is completely consumed first, limiting the amount of product formed

Excess Reactant

The reactant present in greater quantity than needed, remaining after reaction completes

Identification Method:

  1. Convert all reactant masses to moles
  2. Divide by their stoichiometric coefficients
  3. The smallest result indicates the limiting reactant

Example: If 5g H2 reacts with 20g O2, which is limiting?

Solution:

Balanced equation: 2H2 + O2 → 2H2O

Moles H2 = 5g / 2.016g/mol = 2.48 mol → 2.48/2 = 1.24

Moles O2 = 20g / 32g/mol = 0.625 mol → 0.625/1 = 0.625

O2 has smaller ratio → O2 is limiting

5. Solution Stoichiometry

Molarity (M) = moles solute per liter solution

Primary unit for solution concentration in stoichiometry

\[ M = \frac{n}{V} \]

Where:
M = molarity (mol/L)
n = moles of solute
V = volume of solution (L)

Dilution Equation:

\[ M_1V_1 = M_2V_2 \]

Example: What volume of 0.5M HCl is needed to neutralize 25mL of 0.2M NaOH?

Solution:

Reaction: HCl + NaOH → NaCl + H2O (1:1 ratio)

Moles NaOH = 0.2M × 0.025L = 0.005 mol

Moles HCl needed = 0.005 mol

Volume HCl = n/M = 0.005mol / 0.5M = 0.01L (10mL)

Advanced Stoichiometry Concepts

Theoretical Yield

Maximum possible product based on stoichiometry

\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]

Common causes of <100% yield:

  • Incomplete reactions
  • Side reactions
  • Product loss during purification

Gas Stoichiometry

At STP (Standard Temperature and Pressure):

\[ 1 \text{ mole gas} = 22.4 \text{ L at } 0°C, 1 \text{ atm} \]

For non-STP conditions, use the ideal gas law:

\[ PV = nRT \]

Stoichiometry & The Periodic Table

Understanding the periodic table is crucial for stoichiometry:

Atomic Mass

Bottom number in element boxes gives molar mass

Valency

Group number helps predict compound formulas

Periodic Trends

Affects reaction stoichiometry and product formation

Quick Reference:

  • Group 1 elements: +1 charge in compounds
  • Group 2 elements: +2 charge in compounds
  • Group 17 elements: -1 charge in compounds
  • Transition metals: Multiple possible charges

Stoichiometry Mastery Checklist

  • I can convert between mass, moles, and particles
  • I can balance chemical equations
  • I can perform mole-to-mole conversions
  • I can identify limiting reactants
  • I can calculate theoretical and percent yields
  • I can solve solution stoichiometry problems

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