Chemistry Physical - How to Solve: Mole Concept & Stoichiometry (Limiting Reagent, Purity, Percentage Yield) – NEET UG Guide

How to Solve: Mole Concept & Stoichiometry (Limiting Reagent, Purity, Percentage Yield) – NEET UG Guide

Introduction

Mastering mole concept and stoichiometry unlocks 10-15% of NEET Chemistry marks—that’s 40-60 questions in past papers! It’s the backbone of pharmaceutical drug synthesis, industrial chemical production, and even environmental pollution control. If you can solve limiting reagent, purity, and percentage yield problems, you’ll ace NEET, JEE, and board exams—and save hours of last-minute panic.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Mole concept – What a mole is, Avogadro’s number, molar mass.
2. Balanced chemical equations – How to balance reactions and interpret coefficients.
3. Basic percentage calculations – How to find % composition, % yield, and % purity.

If any of these are shaky, stop now and review them first.

KEY TERMS & FORMULAS

1. Mole Concept

• Mole (n) = Amount of substance containing 6.022 × 10²³ particles (Avogadro’s number).
• Molar mass (M) = Mass of 1 mole of a substance (g/mol). Formula: [ n = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} ] MEMORISE THIS

2. Stoichiometry (Mole Ratios)

• Balanced equation coefficients = Mole ratios of reactants and products. Example: 2H₂ + O₂ → 2H₂O means 2 moles H₂ react with 1 mole O₂ to give 2 moles H₂O.

3. Limiting Reagent

• The reactant that runs out first, stopping the reaction. How to find it?
• Calculate moles of each reactant.
• Divide by its coefficient in the balanced equation.
• The smallest value = limiting reagent.

4. Purity of a Substance

• % Purity = (Mass of pure substance / Total mass of sample) × 100 Formula: [ \% \text{Purity} = \left( \frac{\text{Mass of pure substance}}{\text{Total mass}} \right) \times 100 ] MEMORISE THIS

5. Percentage Yield

• % Yield = (Actual yield / Theoretical yield) × 100 Formula: [ \% \text{Yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 ] MEMORISE THIS

STEP-BY-STEP METHOD

Step 1: Write the Balanced Chemical Equation

• If not given, balance the reaction first.
• Example: N₂ + 3H₂ → 2NH₃

Step 2: Convert Given Quantities to Moles

• Use n = mass / molar mass for solids.
• For gases, use n = volume (L) / 22.4 L/mol (at STP).
• For solutions, use n = Molarity × Volume (L).

Step 3: Identify the Limiting Reagent

• Divide moles of each reactant by its coefficient in the balanced equation.
• The smallest value = limiting reagent.

Step 4: Calculate Theoretical Yield

• Use the limiting reagent’s moles and the mole ratio to find product moles.
• Convert product moles to mass (g) using molar mass.

Step 5: Adjust for Purity (If Given)

• If a reactant is impure, multiply its mass by % purity (as a decimal) to get pure mass.
• Example: 10 g of 80% pure CaCO₃ → 8 g pure CaCO₃.

Step 6: Calculate Percentage Yield (If Given)

• Use % Yield = (Actual yield / Theoretical yield) × 100.
• If actual yield is given, plug it in.

WORKED EXAMPLES

Example 1 – Basic (Limiting Reagent)

Question: 2.8 g of N₂ reacts with 1.0 g of H₂ to form NH₃. Find the limiting reagent and the mass of NH₃ produced.

Solution:
1. Balanced equation: N₂ + 3H₂ → 2NH₃
2. Moles of N₂: [ n_{N_2} = \frac{2.8 \text{ g}}{28 \text{ g/mol}} = 0.1 \text{ mol} ]
3. Moles of H₂: [ n_{H_2} = \frac{1.0 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ mol} ]
4. Divide by coefficients: - N₂: 0.1 / 1 = 0.1 - H₂: 0.5 / 3 ≈ 0.167
5. Limiting reagent = N₂ (smaller value)
6. Moles of NH₃ produced (from N₂): [ 0.1 \text{ mol N₂} \times \frac{2 \text{ mol NH₃}}{1 \text{ mol N₂}} = 0.2 \text{ mol NH₃} ]
7. Mass of NH₃: [ 0.2 \text{ mol} \times 17 \text{ g/mol} = 3.4 \text{ g} ]

What we did and why: - Converted masses to moles. - Compared mole ratios to find the limiting reagent. - Used the limiting reagent to calculate product mass.

Example 2 – Medium (Purity & Limiting Reagent)

Question: 10 g of 80% pure CaCO₃ reacts with 50 mL of 1 M HCl. Find the limiting reagent and the mass of CO₂ produced.

Solution:
1. Balanced equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
2. Pure CaCO₃ mass: [ 10 \text{ g} \times 0.8 = 8 \text{ g} ]
3. Moles of CaCO₃: [ n_{CaCO_3} = \frac{8 \text{ g}}{100 \text{ g/mol}} = 0.08 \text{ mol} ]
4. Moles of HCl: [ n_{HCl} = 1 \text{ M} \times 0.05 \text{ L} = 0.05 \text{ mol} ]
5. Divide by coefficients: - CaCO₃: 0.08 / 1 = 0.08 - HCl: 0.05 / 2 = 0.025
6. Limiting reagent = HCl (smaller value)
7. Moles of CO₂ produced (from HCl): [ 0.05 \text{ mol HCl} \times \frac{1 \text{ mol CO₂}}{2 \text{ mol HCl}} = 0.025 \text{ mol CO₂} ]
8. Mass of CO₂: [ 0.025 \text{ mol} \times 44 \text{ g/mol} = 1.1 \text{ g} ]

What we did and why: - Adjusted for purity before calculating moles. - Compared mole ratios to find the limiting reagent. - Used the limiting reagent to find product mass.

Example 3 – Exam-Style (Percentage Yield)

Question: 5 g of impure Zn (90% pure) reacts with excess HCl. If 1.5 L of H₂ gas is collected at STP, what is the percentage yield?

Solution:
1. Balanced equation: Zn + 2HCl → ZnCl₂ + H₂
2. Pure Zn mass: [ 5 \text{ g} \times 0.9 = 4.5 \text{ g} ]
3. Moles of Zn: [ n_{Zn} = \frac{4.5 \text{ g}}{65 \text{ g/mol}} ≈ 0.0692 \text{ mol} ]
4. Theoretical moles of H₂ (1:1 ratio): [ 0.0692 \text{ mol} ]
5. Theoretical volume of H₂ (at STP): [ 0.0692 \text{ mol} \times 22.4 \text{ L/mol} ≈ 1.55 \text{ L} ]
6. Actual yield = 1.5 L
7. % Yield: [ \left( \frac{1.5 \text{ L}}{1.55 \text{ L}} \right) \times 100 ≈ 96.77\% ]

What we did and why: - Adjusted for purity first. - Calculated theoretical yield using mole ratios. - Compared actual vs. theoretical to find % yield.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for mole concept and stoichiometry in NEET!

1. Always start with a balanced equation—no shortcuts.
2. Convert everything to moles first—mass to moles, volume to moles, molarity to moles.
3. Find the limiting reagent by dividing moles by coefficients—smallest value wins.
4. Adjust for purity—multiply mass by % purity (as a decimal).
5. Calculate theoretical yield from the limiting reagent.
6. Percentage yield = (Actual / Theoretical) × 100.
7. Watch out for traps—excess reactants, % impurity, unit mismatches.

You’ve got this! Now go solve those problems like a pro."