CUET UG Chemistry: Physical Chemistry - Mole Concept and Stoichiometry, Limiting Reagent, Percentage Yield

Must-Know

• The mole is the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12 (verify from NCERT).
• 1 mole = 6.022 × 10²³ particles (Avogadro’s number), used to convert between atomic and macroscopic scales.
• Molar mass of a substance (in g/mol) is numerically equal to its atomic/molecular mass in u; e.g., O? = 32 g/mol.
• Number of moles = given mass (g) / molar mass (g/mol); e.g., 4 g of NaOH = 4/40 = 0.1 mol.
• For gases at STP (0°C, 1 atm), 1 mole occupies 22.4 L; e.g., 2 mol O? occupies 44.8 L at STP.
• Stoichiometry uses balanced chemical equations to determine quantitative relationships between reactants and products.
• In the reaction N? + 3H?-2NH?, 1 mole N? reacts with 3 moles H? to produce 2 moles NH?.
• The limiting reagent is the reactant completely consumed first, determining the maximum amount of product formed.
• In a reaction where 2 mol H? and 1 mol O? react to form H?O (2H? + O?-2H?O), H? is limiting if only 1.5 mol H? is present.
• To identify limiting reagent: calculate mole ratio from equation and compare with actual ratio; smaller ratio indicates limiting reagent.
• Theoretical yield is the maximum amount of product calculated from stoichiometry based on limiting reagent.
• Actual yield is the experimentally measured amount of product, always-theoretical yield.
• Percentage yield = (actual yield / theoretical yield) × 100; e.g., if theoretical = 10 g, actual = 8 g, % yield = 80%.
• In the reaction 2Mg + O?-2MgO, if 48 g Mg (2 mol) reacts with 32 g O? (1 mol), Mg and O? are in stoichiometric ratio; neither is limiting.
• If 24 g Mg (1 mol) reacts with 32 g O? (1 mol), Mg is limiting because 1 mol Mg requires only 0.5 mol O?.
• Excess reagent is the reactant not completely used up; amount left = initial – consumed (based on limiting reagent).
• For 2A + B-C, if 5 mol A and 2 mol B are given, B is limiting (required A:B = 2:1; available = 5:2 = 2.5:1 > 2:1).
• Mass-mass stoichiometry involves: mass A-moles A-moles B (via stoichiometric ratio)-mass B.
• In combustion of methane: CH? + 2O?-CO? + 2H?O, 16 g CH? produces 44 g CO? theoretically.
• Percentage purity = (mass of pure substance / mass of impure sample) × 100; used when reactants are not 100% pure.

Difficulty Level

Intermediate — requires conceptual clarity in identifying limiting reagents and multi-step calculations involving mole conversions and yield.

Common CUET Traps

• Trap: Assuming the reactant with smaller mass is always limiting. Avoid: Compare mole ratios using stoichiometric coefficients, not mass.
• Trap: Using actual yield in stoichiometric calculations instead of theoretical yield. Avoid: Theoretical yield comes from limiting reagent; actual yield is only for % yield.
• Trap: Forgetting to balance the chemical equation before stoichiometry. Avoid: Always start with a balanced equation; coefficients define mole ratios.

Practice MCQs

1. Question: How many moles of water are produced when 2 moles of hydrogen gas react with 1 mole of oxygen gas?
   A. 1 mol
   B. 2 mol
   C. 3 mol
   D. 4 mol
   Answer: B
   Explanation: From 2H? + O?-2H?O, 2 mol H? produces 2 mol H?O.
   Why others fail: Option A assumes 1:1 ratio without checking stoichiometry.

2. Question: What is the mass of ammonia produced when 14 g N? reacts with excess H (N? + 3H?-2NH?)
   A. 17 g
   B. 34 g
   C. 8.5 g
   D. 3.4 g
   Answer: A
   Explanation: 14 g N? = 0.5 mol-produces 1 mol NH? = 17 g.
   Why others fail: Option B doubles the correct answer, assuming 1 mol N? gives 2 mol NH? but misusing mass.

3. Question: 3 g of H? reacts with 28 g of N?. Which is the limiting reagent? (N? + 3H?-2NH?)
   A. N?
   B. H?
   C. Both are limiting
   D. Neither is limiting
   Answer: B
   Explanation: Moles H? = 1.5, moles N? = 1; required H? for 1 mol N? = 3 mol, but only 1.5 mol H?-H? limiting.
   Why others fail: Option A is tempting if comparing masses directly (3 g vs 28 g).

4. Question: A reaction has a theoretical yield of 50 g, but only 40 g is obtained. What is the percentage yield?
   A. 80%
   B. 90%
   C. 75%
   D. 85%
   Answer: A
   Explanation: (40/50) × 100 = 80%.
   Why others fail: Option C is common error from incorrect division (40/50 = 0.75).

5. Question: Calcium carbonate decomposes: CaCO?-CaO + CO?. If 50 g of impure CaCO? (80% pure) is heated, what is the mass of CO? produced?
   A. 17.6 g
   B. 22 g
   C. 13.2 g
   D. 8.8 g
   Answer: A
   Explanation: Pure CaCO? = 40 g = 0.4 mol-produces 0.4 mol CO? = 0.4 × 44 = 17.6 g.
   Why others fail: Option B assumes 50 g pure CaCO?-1 mol CO? without adjusting for purity.

Last-Minute Revision

• Limiting reagent determines theoretical yield.
• % yield = (actual / theoretical) × 100 — never exceeds 100%.
• STP: 1 mol gas = 22.4 L at 0°C and 1 atm.
• Avogadro’s number = 6.022 × 10²³ mol?¹.
• Molar mass in g/mol = molecular mass in u.
• In 2A + B-C, mole ratio A:B is 2:1.
• Excess reagent remains after reaction ends.
• Always balance chemical equations before stoichiometry.
• Mass-moles-mole ratio-moles of product-mass of product.
• For 2H? + O?-2H?O, 2 g H? requires 16 g O?.
• 1 mole of O? = 32 g = 22.4 L at STP.
• Percentage purity affects reactant availability.
• Theoretical yield is calculated, not measured.
• Actual yield-theoretical yield.
• Mnemonic: "LEO the lion says GER" – Loss of Electrons = Oxidation, Gain = Reduction (not directly on topic but high-yield redox reminder).
• In N? + 3H?-2NH?, 28 g N? produces 34 g NH?.