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ChemistryMediumClass 11

Stoichiometry — Mole ratios from equations

Mole Concept

4

JEE Qs

8%

Hard

60

min

Master the art of balancing chemical equations, as it is the foundational step for all stoichiometry problems and a common source of error.

🧮 Key Formulas

n = m / M (moles = mass / molar mass)
n = N / N_A (moles = number of particles / Avogadro's number, where N_A is Avogadro's number)
For a balanced chemical reaction aA + bB → cC + dD, the mole ratios are given by: n_A / a = n_B / b = n_C / c = n_D / d

✅ Key Points for JEE

  • 1Always start by writing a balanced chemical equation, as incorrect balancing leads to completely wrong mole ratios and subsequent calculations.
  • 2The stoichiometric coefficients in a balanced equation represent the exact mole ratios in which reactants combine and products are formed.
  • 3All quantitative calculations involving chemical reactions must be performed by converting given quantities (mass, volume, number of particles) into moles, using mole ratios, and then converting back to the desired quantity.
  • 4For gaseous reactions at constant temperature and pressure, the stoichiometric coefficients also represent the volume ratios of reactants and products.

⚠️ Common Mistakes

  • Failing to balance the chemical equation correctly before using coefficients as mole ratios.
  • Using mass ratios directly from stoichiometric coefficients instead of converting to mole ratios first.
  • Incorrectly converting between mass and moles, or number of particles and moles, leading to errors in the initial mole calculation.
  • Confusing volume ratios with mole ratios for non-gaseous substances.

📝 Practice Questions

See all

Q71.Xg of benzoic acid on reaction with aq NaHCO3 released CO2 that occupied 11.2 L volume at STP. X is _____ g.

2025·NumericalMedium

Q60.The elemental composition of a compound is 54.2%C, 9.2%H and 36.6%O. If the molar mass of the compound is 132 g mol−1 , the molecular formula of the compound is : [Given : The relative atomic mass of C : H : O = 12 : 1 : 16 ] (1) C4H9O3 (2) C6H12O6 (3) C4H8O2 (4) C6H12O3

2025·MCQEasy

Q67.Match the LIST-I with LIST-II Choose the correct answer from the options given below: (1) A-II, B-I, C-III, D-IV (2) A-II, B-III, C-I, D-IV (3) A-IV, B-I, C-III, D-II (4) A-IV, B-III, C-I, D-II

2025·MCQEasy

Q73.When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is_______ - (Nearest integer) Given : Molar mass of Al is 27.0 g mol−1 Molar mass of O is 16.0 g mol−1

2025·NumericalMedium

Q72.Consider the following reaction occurring in the blast furnace: Fe3O4( s) + 4CO(g) →3Fe(l) + 4CO2( g) ' x ' kg of iron is produced when 2.32 × 103 kgFe3O4 and 2.8 × 102 kgCO are brought together in the furnace. The value of ' x ' is _____. (nearest integer) Given: molar mass of Fe3O4 = 232 g mol−1 molar mass of CO = 28 g mol−1 molar mass of Fe = 56 g mol−1} Q73. 37.8 g N2O5 was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K 2 N2O5( g) ⇌2 N2O4( g) + O2( g) The total pressure at equilibrium was found to be 18.65 bar. Then, Kp = ______ ×10−2 [nearest integer] Assume N2O5 to behave ideally under these conditions. Given: R = 0.082 bar Lmol−1 K−1

2025·NumericalMedium

Q31.Combustion of glucose (C6H12O6) produces CO2 and water. The amount of oxygen (in g) required for the complete combustion of 900 g of glucose is : [Molar mass of glucose in gmol−1 = 180 ] (1) 480 (2) 800 (3) 960 (4) 32

2024·MCQEasy

NCERT Chapters

  • Class 11 Chemistry Ch 1: Some Basic Concepts of Chemistry