## Stoichiometry CBSE Class 11 Chemistry

**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 CO

_{2}that can be obtained by heating 50 gm of limestone.

�

�

**Solution**

�Step I - Write a Balanced Equation

�CaCO

_{3}----> CaO + CO

_{2}

�

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

�CaCO3 ----> CaO + CO

_{2}

�

�

**Method I - MOLE METHOD**

�Number of moles of 50 gm of CaCO

_{3}= 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 CO

_{2 }= Moles x Molecular Mass

�= 0.5 x 44

�= 22 gm

�

�

**Method II - FACTOR METHOD**

�From equation we may write as

�100 gm of CaCO

_{3}gives 44 gm of CO

_{2}

�1 gm of CaCO

_{3}will give 44/100 gm of CO

_{2}

�50 gm of CaCO

_{3}will give 50 x 44 / 100 gm of CO

_{2}

�= 22 gm of CO

_{2}

�

�

**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.

dm3 volume at S.T.P

�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 CO

_{2}gas produced at S.T.P by combustion of 20 gm of CH

_{4}.

�

�

**Solution**

�Step I - Write a Balanced Equation

�CH

_{4}+ 2 O

_{2}----> CO

_{2}+ 2 H

_{2}O

�

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

�CH

_{4}+ 2 O

_{2}----> CO

_{2}+ 2 H

_{2}O

�

�

**Method I - MOLE METHOD**

�Convert the given mass of CH4 in moles

�Number of moles of CH

_{4}= Given Mass of CH4 / Molar Mass of CH4

�From Equation

�1 mole of CH4 gives 1 moles of CO

_{2}

�1.25 mole of CH

_{4}will give 1.25 mole of CO

_{2}

�No. of moles of CO

_{2}obtained = 1.25

�But 1 mole of CO

_{2}at S.T.P occupies 22.4 dm

^{3 }

�1.25 mole of CO

_{2}at S.T.P occupies 22.4 x 1.25

�= 28 dm

^{3 }

�

�

**Method II - FACTOR METHOD**

�Molecular mass of CH

_{4}= 16

�Molecular mass of CO

_{2 }= 44

�According to the equation

�16 gm of CH

_{4}gives 44 gm of CO

_{2 }

�1 gm of CH

_{4}will give 44/16 gm of CO

_{2 }�

�20 gm of CH

_{4}will give 20 x 44/16 gm of CO

_{2 }

�= 55 gm of CO

_{2 }

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

^{3}

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

^{3 }

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

�= 28 dm

^{3 }

�

�

**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.

conditions of temperature & pressure

�Consider this equation

�CH

_{4}+ 2 O

_{2}----> CO

_{2}+ 2 H

_{2}O

�In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to

give one volume of CO

_{2}and two volumes of H

_{2}O

�

�

**Examples**

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

_{2}H

_{4}

(ethylene)?

�

�

**Solution**

�Step I - Write a Balanced Equation

�C

_{2}H

_{4}+ 3 O

_{2}----> 2 CO

_{2}+ 2 H

_{2}O

�

�

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

�C

_{2}H

_{4}+ 3 O

_{2}----> 2 CO

_{2}+ 2 H

_{2}O

�

�According to Equation

�1 dm

^{3}of C

_{2}H

_{4}requires 3 dm

^{3}of O

_{2}

�500 dm

^{3}of C

_{2}H

_{4}requires 3 x 500 dm

^{3}of O

_{2}

�= 1500 dm

^{3}of O

_{2 }

�

**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.

�N

_{2}+ 3 H

_{2}----> 2 NH

_{3}

�

�

**Solution**

�In this problem moles of N

_{2}and H

_{2}are as follows

�Moles of N

_{2}= Mass of N

_{2}/ Mol. Mass of N

_{2}

�= 50 / 28

�= 1.79

�Moles of H2 = Mass of H

_{2}/ Mol. Mass of H

_{2}

�= 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.

�N

_{2}+ 3 H

_{2}----> 2 NH

_{3 }

�

�1 mole of N

_{2}gives 2 moles of NH

_{3}

�1.79 mole of N

_{2}gives 2 x 1.79 moles of NH

_{3}

�= 3.58 moles of NH

_{3 }

�

�Mass of NH

_{3}= Moles of NH

_{3}x Mol. Mass

�= 3.58 x 17

�= 60.86 gm of NH

_{3}