Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
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CHAPTER 7: EQUILIBRIUM
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
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CHAPTER 10: S-BLOCK ELEMENTS
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
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CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
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Class 11th Chemistry Online Class For 100% Result
About Lesson

Stoichiometry

  • The word ‘stoichiometry’ is derived from two Greek words – stoicheion (meaning element) and metron (meaning measure).
  • Stoichiometry deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction.
  • This is done using balance chemical equation.

Chemical Reaction

  • A chemical reaction takes place whenever a chemical change occurs.
    • For example: – Magnesium (Mg) + Oxygen (O) à Magnesium oxide (MgO).
  • Reactants are the substances which undergo chemical change in the reaction.
  • Products are the new substances, formed during the reaction.
    • For example: – Magnesium (Mg) + Oxygen (O) à Magnesium oxide (MgO). In this equation Magnesium and Oxygen are reactants and Magnesium oxide is the product formed.

Balance Chemical Reaction

  • According to law of conservation of mass; mass can neither be created nor be destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants.
  • Number of atoms on each element remains the same, before and after a chemical reaction.
    • For example:- (a) 2Mg + O2 à 2MgO
    • (b) Zn + H2SO4 à  ZnSO4 + H2

Balance Chemical equation in Stoichiometry

  • CH 4(g) + 2O(g) à CO(g) + 2H2O (g).
  • The above reaction gives the information as follows:-
    • One mole of CH(g) reacts with two moles of O(g) to give one mole of CO(g) and two moles of H2O (g).
    • One molecule of CH(g) reacts with 2 molecules of O(g) to give one molecule of CO(g) and 2 molecules of H2O (g).
    • 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2×18 g of H2O (g).

 Problem:-

Calculate the amount of water (g) produced by the combustion of 16 g of methane.

Answer:-

The balanced equation for combustion of methane is:

CH(g) + 2O2 (g) à CO(g) +2H2O (g)

(i) 16 g of CH4 corresponds to one mole.

(ii) From the above equation, 1 mol of CH4 (g) gives 2 mol of H2O (g).

2 mol of water (H2O) = 2 × (2+16) = 2 × 18 = 36 g

1 mol H2O = 18 g H2O ⇒ (18g H2O)/ (1 mol H2O) = 1

Hence 2 mol H2O × (18g H2O)/ (1 mol H2O)

= 2 × 18 g H2O = 36 g H2O

Limiting Reagent

  • In a chemical reaction, reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is, therefore, called the limiting reagent.

Problem:-

50.0 kg of N(g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g).

Calculate the NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.

Answer:-

A balanced equation for the above reaction is written as follows:

Calculation of moles:

N2 (g) + 3H2 (g) ⇌2NH3 g

moles of N2

= 50.0 kg N2 × (1000 g N)/ (1 kgN2) x (1molN2)/ (28.0g N2)

= 17.86×102 mol

moles of H2

= 10.00 kg H2 × (1000g H2)/ (1kg H2)/ (1molH2)/ (2.016g H2)

= 4.96×103 mol

According to the above equation, 1 mol

N2 (g) requires 3 mol H2 (g), for the reaction.

Hence, for 17.86×102 mol of N2, the moles of H2 (g) required would be

17.86×102 mol N2 × (3 molH(g))/ (1mol N2 (g)

= 5.36 ×103 mol H2

But we have only 4.96×103 mol H2. Hence, dihydrogen is the limiting reagent in this case. So NH(g) would be formed only from that amount of available dihydrogen i.e., 4.96 × 103 mol

Since 3 mol H(g) gives 2 mol NH(g)

(4.96×103 mol H(g)) × (2molNH3 (g))/ (3molH2 (g))

= 3.30×103 mol NH3 (g)

=3.30×103 mol NH3 (g) is obtained.

If they are to be converted to grams, it is done as follows:

1 mol NH3 (g) = 17.0 g NH(g)

3.30×103 mol NH3 (g) × (17.0 g NH3 (g))/ (1mol NH3 (g))

= 3.30×103×17g NH3 (g)

= 56.1×103g NH3

= 56.1 kg NH3

Problem:-

Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer:

The balanced reaction of combustion of carbon can be written as:

(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.

(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant.

Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

Problem:-

In a reaction: – A + B2 → AB2

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv)  5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

Answer:-

A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mole of A reacts with 1 mole of B. Thus, 2 mole of A will react with only 2 mole of B. As a result, 1 mole of A will not be consumed. Hence, A is the limiting reagent.

(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mole of atom A combines with 1 mole of molecule B. Thus, 2.5 mole of B will combine with only 2.5 mole of A. As a result, 2.5 mole of A will be left as such. Hence, B is the limiting reagent.

 (v) According to the reaction, 1 mole of atom A combines with 1 mole of molecule B. Thus, 2.5 mole of A will combine with only 2.5 mole of B and the remaining 2.5 mole of B will be left as such. Hence, A is the limiting reagent.

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