Limiting Reagent — Finding limiting reagent
Mole Concept
4
JEE Qs
8%
Hard
75
min
Always balance the chemical equation first and work with moles, not masses, to accurately identify the limiting reagent and base all further calculations on it.
🧮 Key Formulas
✅ Key Points for JEE
- 1Always begin by writing and balancing the chemical equation to ensure correct stoichiometric ratios.
- 2The limiting reagent (or reactant) is the reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed.
- 3To identify the limiting reagent, calculate the number of moles for each reactant. Then, divide the moles of each reactant by its respective stoichiometric coefficient from the balanced equation. The reactant yielding the smallest ratio is the limiting reagent.
- 4Alternatively, calculate the theoretical amount of product formed if each reactant were completely consumed. The reactant that produces the least amount of product is the limiting reagent.
- 5All subsequent calculations regarding product formation or the amount of excess reagent remaining must be based *solely* on the limiting reagent.
⚠️ Common Mistakes
- ✕Not balancing the chemical equation correctly, leading to incorrect stoichiometric ratios.
- ✕Comparing the initial moles or masses of reactants directly without considering their stoichiometric coefficients from the balanced equation.
- ✕Mistaking the reactant with the smallest initial quantity (moles or mass) as the limiting reagent without proper calculation.
- ✕Using the excess reagent instead of the limiting reagent for calculating the amount of product formed or other reaction outcomes.
📝 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