Class 11 CBSE Ch 1 Some basic concept of chemistry

3.  Chemical Reactions
 Stoichiometry

4. One can calculate the empirical formula from percentage composition
In the last part we have seen about empirical formula and
solved some problems over it.

5.  To construct a balanced chemical equation
 Write the reactants on the left side
 Products on the right side
 Balance the equation using the smallest possible
coefficient
 Never change a subscript of a molecular formula
 Symbol indicating physical state for each of the reactant
and product are shown in the chemical equations as
 Gas (g)
 Liquid (l)
 Solid (s)
 Aqueous (aq)

6.  Coefficient in a chemical equation represents the relative numbers of molecules in a
reaction.
Quantitative information from balanced equation

7. The word ‘stoichiometry’ is derived from two Greek words —
stoicheion (meaning, element) and metron (meaning, measure).
 Stoichiometry is the calculation of reactants and products in
chemical reactions.
 Stoichiometry is founded on the law of conservation of mass where
the total mass of the reactants equals the total mass of the products,
leading to the insight that the relations among quantities of reactants
and products typically form a ratio of positive integers.
 This means that if the amounts of the separate reactants are known,
then the amount of the product can be calculated. Conversely, if one
reactant has a known quantity and the quantity of the products can be
empirically determined, then the amount of the other reactants can
also be calculated.

8. Example
Here, one molecule of methane reacts with two molecules of oxygen gas to yield one
molecule of carbon dioxide and two molecules of water. This particular chemical
equation is an example of complete combustion.
• One mole of CH4(g) reacts with two moles of O2(g) to give one mole of CO2(g) and two moles
of H2O(g)
• One molecule of CH4(g) reacts with 2 molecules of O2(g) to give one molecule of CO2(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). From
these relationships, the given data can be interconverted as follows:
Mass ⇌ Moles ⇌ No. of molecules

9.  Stoichiometry measures these quantitative relationships, and is used to
determine the amount of products and reactants that are produced or needed in
a given reaction.
 Describing the quantitative relationships among substances as they participate in
chemical reactions is known as reaction stoichiometry. In the example above,
reaction stoichiometry measures the relationship between the quantities of
methane and oxygen that react to form carbon dioxide and water.
 A stoichiometric amount or stoichiometric ratio of a reagent is the
optimum amount or ratio where, assuming that the reaction proceeds to
completion:
1.All of the reagent is consumed
2.There is no deficiency of the reagent
3.There is no excess of the reagent.

10.  Stoichiometry rests upon the very basic laws that help to understand it better,
i.e., law of conservation of mass, the law of definite proportions (i.e., the law of
constant composition), the law of multiple proportions and the law of reciprocal
proportions.
 In general, chemical reactions combine in definite ratios of chemicals. Since
chemical reactions can neither create nor destroy matter, nor transmute one
element into another, the amount of each element must be the same throughout
the overall reaction.

11. Step 1: Write a balanced chemical equation
Step 2: If you’re given m or N of a substance, Convert it to the number of
moles.
Step 3: Calculate the number of moles of the required substance based
on the number of moles of the given substance (using appropriate
mole ratio)
Step 4: Convert the number of moles of the required substance to mass
or number of particles, as directed by the question.

13. Another example showing the coefficients in a balanced
chemical equation specify the relative amount of each of
the substances involved in the reaction

14. How to balance a chemical reaction?

15. The following Stoichiometry Road Map gives a summary of how to use stoichiometry
to calculate moles, masses, volumes and particles in a chemical reaction with limiting
and excess reactants.

16.  The limiting reagent (or limiting reactant or limiting agent) in a chemical
reaction is the substance that is totally consumed when the chemical reaction
is completed.
 The amount of product formed is limited by this reagent, since the reaction
cannot continue without it.
 If one or more other reagents are present in excess of the quantities required
to react with the limiting reagent, they are described as excess reagents or
excess reactants (xs).
 The limiting reagent must be identified in order to calculate the percentage
yield of a reaction since the theoretical yield is defined as the amount of
product obtained when the limiting reagent reacts completely.
 Given is the general example, which describes how to find limiting reagent
and evaluate the excess quantities of other reagents.

17.  Assuming the hot dogs and buns combine in a one-to-one ratio, we will be
limited by the number of hot dog buns we have since we will run out of buns
first.
 In this less than ideal situation, we would call the hot dog buns the limiting
reagent or limiting reactant.
 The maximum amount of product that can be produced is called the theoretical
yield. In the case of the hot dogs and hot dog buns, our theoretical yield is four
complete hot dogs, since we have four hot dog buns.

18. General example and Chemical equation showing limiting and
excess reagent

19. There are several equivalent ways to identify the
limiting reagent and evaluate the excess quantities of
other reagents in a chemical reaction.
That we will learn in next part.