Mole Concept + Stoichiometry
Mole Concept
40
JEE Qs
5%
Hard
100
min
Master unit conversions and balancing chemical equations; always recheck your limiting reagent calculations as they are the pivot for most stoichiometry problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1The mole is a fundamental counting unit that connects macroscopic quantities (mass, volume) to microscopic entities (atoms, molecules, ions).
- 2A balanced chemical equation is indispensable for all stoichiometric calculations, as it provides the exact mole ratios between reactants and products.
- 3Always identify the limiting reagent in reactions with multiple reactants; all product calculations must be based on the amount of the limiting reagent.
- 4Pay meticulous attention to units (g vs kg, mL vs L, STP vs NTP) and perform accurate conversions to avoid calculation errors.
- 5Understanding and correctly applying various concentration terms (Molarity, Molality, Mole fraction) is vital for solution-based problems.
⚠️ Common Mistakes
- ✕Failing to balance the chemical equation correctly before attempting any stoichiometric calculation.
- ✕Not identifying or incorrectly identifying the limiting reagent, leading to wrong theoretical yield or product amounts.
- ✕Confusing Molarity (moles/volume of solution) with Molality (moles/mass of solvent) or using incorrect units for volume/mass.
- ✕Incorrectly using gas volume at STP/NTP (e.g., using 22.4 L for NTP or 22.7 L for STP, or for conditions other than standard).
- ✕Errors in basic unit conversions (e.g., mg to g, cm^3 to L, kg to g) which propagate through calculations.
📝 Practice Questions
See allQ71.Xg of benzoic acid on reaction with aq NaHCO3 released CO2 that occupied 11.2 L volume at STP. X is _____ g.
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
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
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
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
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
NCERT Chapters
- Class 11 Chemistry Ch 1: Some Basic Concepts of Chemistry