Stoichiometry CBSE Class 11 Chemistry

9/24/2015 CBSE

Stoichiometry



(Calculation Based On Chemical Equations)



Definition

The study of relationship between the amount of reactant and the products in
chemical reactions as given by chemical equations is called stoichiometry.



�In this study we always use a balanced chemical equation because a balanced
chemical equation tells us the exact mass ratio of the reactants and products in
the chemical reaction.

�There are three relationships involved for the stoichiometric calculations from
the balanced chemical equations which are

�1. Mass - Mass Relationship

�2. Mass - Volume Relationship

�3. Volume - Volume Relationship



Mass - Mass Relationship

�In this relationship we can determine the unknown mass of a reactant or product
from a given mass of teh substance involved in the chemical reaction by using a
balanced chemical equation.



Example

�Calculate the mass of CO2 that can be obtained by heating 50 gm of limestone.




Solution

�Step I - Write a Balanced Equation

�CaCO3 ----> CaO + CO2



�Step II - Write Down The Molecular Masses And Moles Of Reactant & Product

�CaCO3 ----> CaO + CO2



Method I - MOLE METHOD

�Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole

�According to equation

�1 mole of CaCO3 gives 1 mole of CO2

�0.5 mole of CaCO3 will give 0.5 mole of CO2

�Mass of CO2 = Moles x Molecular Mass

�= 0.5 x 44

�= 22 gm



Method II - FACTOR METHOD

�From equation we may write as

�100 gm of CaCO3 gives 44 gm of CO2

�1 gm of CaCO3 will give 44/100 gm of CO2

�50 gm of CaCO3 will give 50 x 44 / 100 gm of CO2

�= 22 gm of CO2



Mass - Volume Relationship

�The major quantities of gases can be expressed in terms of volume as well as
masses. According to Avogardro One gm mole of any gas always occupies 22.4
dm3 volume at S.T.P
. So this law is applied in mass-volume relationship.

�This relationship is useful in determining the unknown mass or volume of
reactant or product by using a given mass or volume of some substance in a
chemical reaction.



Example

�Calculate the volume of CO2 gas produced at S.T.P by combustion of 20 gm of CH4.




Solution

�Step I - Write a Balanced Equation

�CH4 + 2 O2 ----> CO2 + 2 H2O



�Step II - Write Down The Molecular Masses And Moles Of Reactant & Product

�CH4 + 2 O2 ----> CO2 + 2 H2O



Method I - MOLE METHOD

�Convert the given mass of CH4 in moles

�Number of moles of CH4 = Given Mass of CH4 / Molar Mass of CH4

�From Equation

�1 mole of CH4 gives 1 moles of CO2

�1.25 mole of CH4 will give 1.25 mole of CO2

�No. of moles of CO2 obtained = 1.25

�But 1 mole of CO2 at S.T.P occupies 22.4 dm3

�1.25 mole of CO2 at S.T.P occupies 22.4 x 1.25

�= 28 dm3



Method II - FACTOR METHOD

�Molecular mass of CH4 = 16

�Molecular mass of CO2 = 44

�According to the equation

�16 gm of CH4 gives 44 gm of CO2

�1 gm of CH4 will give 44/16 gm of CO2

�20 gm of CH4 will give 20 x 44/16 gm of CO2

�= 55 gm of CO2

�44 gm of CO2 at S.T.P occupy a volume 22.4 dm3

�1 gm of CO2 at S.T.P occupy a volume 22.4/44 dm3

�55 gm of CO2 at S.T.P occupy a volume 55 x 22.4/44

�= 28 dm3



Volume - Volume Relationship

�This relationship determine the unknown volumes of reactants or products from a
known volume of other gas.

�This relationship is based on Gay-Lussac's law of combining volume which states
that gases react in the ratio of small whole number by volume under similar
conditions of temperature & pressure
.

�Consider this equation

�CH4 + 2 O2 ----> CO2 + 2 H2O

�In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to
give one volume of CO2 and two volumes of H2O



Examples

�What volume of O2 at S.T.P is required to burn 500 litres (dm3) of C2H4
(ethylene)?



Solution

�Step I - Write a Balanced Equation

�C2H4 + 3 O2 ----> 2 CO2 + 2 H2O



Step II - Write Down The Moles And Volume Of Reactant & Product

�C2H4 + 3 O2 ----> 2 CO2 + 2 H2O



�According to Equation

�1 dm3 of C2H4 requires 3 dm3 of O2


�500 dm3 of C2H4 requires 3 x 500 dm3 of O2


�= 1500 dm3 of O2

Limiting Reactant



�In stoichiometry when more than one reactant is involved in a chemical reaction,
it is not so simple to get actual result of the stoichiometric problem by making
relationship between any one of the reactant and product, which are involved in
the chemical reaction. As we know that when any one of the reactant is
completely used or consumed the reaction is stopped no matter the other
reactants are present in very large quantity. This reactant which is totally
consumed during the chemical reaction due to which the reaction is stopped is
called limiting reactant.

�Limiting reactant help us in calculating the actual amount of product formed
during the chemical reaction. To understand the concept the limiting reactant
consider the following calculation.



Problem

�We are provided 50 gm of H2 and 50 gm of N2. Calculate how many gm of NH3 will
be formed when the reaction is irreversible.

�The equation for the reaction is as follows.

�N2 + 3 H2 ----> 2 NH3



Solution

�In this problem moles of N2 and H2 are as follows

�Moles of N2 = Mass of N2 / Mol. Mass of N2

�= 50 / 28

�= 1.79

�Moles of H2 = Mass of H2 / Mol. Mass of H2

�= 50 / 2

�= 25

�So, the provided moles for the reaction are

�nitrogen = 1.79 moles and hydrogen = 25 moles

�But in the equation of the process 1 mole of nitrogen require 3 mole of
hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3
moles of hydrogen i.e. 5.37 moles although 25 moles of H2 are provided but when
nitrogen is consumed the reaction will be stopped and the remaining hydrogen is
useless for the reaction so in this problem N2 is a limiting reactant by which
we can calculate the actual amount of product formed during the reaction.

�N2 + 3 H2 ----> 2 NH3



�1 mole of N2 gives 2 moles of NH3

�1.79 mole of N2 gives 2 x 1.79 moles of NH3

�= 3.58 moles of NH3



�Mass of NH3 = Moles of NH3 x Mol. Mass

�= 3.58 x 17

�= 60.86 gm of NH3

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