POAC — Principle of Atom Conservation
Mole Concept
4
JEE Qs
8%
Hard
60
min
Always clearly identify the specific atom you are conserving and meticulously count its presence in both reactants and products to avoid errors in POAC applications.
🧮 Key Formulas
✅ Key Points for JEE
- 1POAC is a direct application of the Law of Conservation of Mass to individual atoms, stating that the number of atoms of each specific element remains constant throughout a chemical reaction.
- 2It is especially powerful for stoichiometric calculations when the balanced chemical equation is unknown, difficult to balance, or involves multi-step reactions where intermediate products are not isolated or specified.
- 3To apply POAC, identify the specific atom whose quantity is to be conserved, and account for its presence (number of atoms per molecule/formula unit) in all relevant reactant(s) and product(s).
- 4POAC relates the amount of a specific element in the starting material to the amount of the same element in the final product(s), bypassing the need for a full balanced equation.
⚠️ Common Mistakes
- ✕Applying POAC to entire compounds or molecules instead of a specific, individual atom.
- ✕Incorrectly counting the number of atoms of the conserved element within the chemical formula of reactants or products.
- ✕Failing to account for all reactants or products that contain the conserved atom.
- ✕Confusing POAC with limiting reagent concepts when both are applicable in a problem.
📝 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