Stoichiometry Equations

Stoichiometry Equations

Stoichiometry

Stoichiometry is the calculation of the quantities of reactants and products involved in the chemical reaction. It is the study of the relationship between the number of mole of the reactants and products of a chemical reaction. A stoichiometric equation is a short scientific representation of a chemical reaction.

Rules for writing stoichiometric equations

i.In order to write the stoichiometric equation correctly, we must know the reacting substances, all the products formed and their chemical formula.

ii.The formulae of the reactant must be written on the left side of arrow with a positive sign between them.

iii.The formulae of the products formed are written on the right side of the arrow mark. If there is more than one product, a positive sign is placed between them. The equation thus obtained is called skeleton equation. For example, the Chemical reaction between Barium chloride and sodium sulphate producing BaSO4 and NaCl is represented by the equation as

BaCl₂ + Na₂SO₄ ® BaSO₄ + NaCl

This skeleton equation itself is a balanced one. But in many cases the skeleton equation is not a balanced one.

For example, the decomposition of Lead Nitrate giving Lead oxide, NO2 and oxygen. The skeletal equation for this reaction is

Pb(NO3)2 ® PbO + NO2 + O2

iv.In the skeleton equation, the numbers and kinds of particles present on both sides of the arrow are not equal.

 During balancing the equation, the formulae of substances should not be altered, but the number of molecules with it only be suitably changed.

vi. Important conditions such as temperature, pressure, catalyst etc., may be noted above (or) below the arrow of the equation.

 vii.An upward arrow (-) is placed on the right side of the formula of a gaseous product and a downward arrow (¯) on the right side of the formulae of a precipitated product.

 viii.All the reactants and products should be written as molecules including the elements like hydrogen, oxygen, nitrogen, fluorine chlorine, bromine and iodine as H2, O2, N2, F2, Cl2, Br2 and I2

Balancing chemical equation in its molecular form

A chemical equation is called balanced equation only when the numbers and kinds of molecules present on both sides are equal. The several steps involved in balancing chemical equation are discussed below.

Example 1

Hydrogen combines with bromine giving HBr

H₂ + Br₂   -> HBr

This is the skeletal equation. The number of atoms of hydrogen on the left side is two but on the right side it is one. So the number of molecules of HBr is to be multiplied by two. Then the equation becomes

H2 + Br2    -> 2HBr

This is the balanced (or) stoichiometric equation.

Example 2

Potassium permanganate reacts with HCl to give KCl and other products. The skeletal equation is

KMnO₄ + HCl   -> KCl + MnCl₂ + H₂O + Cl₂

If an element is present only one substance in the left hand side of the equation and if the same element is present only one of the substances in the right side, it may be taken up first while balancing the equation.

According to the above rule, the balancing of the equation may be started with respect to K, Mn, O (or) H but not with Cl.

There are   

L.H.S.,  R.H.S

K = 1, 1

Mn = 1,       1

O = 4, 1

So the equation becomes

KMnO4 + HCl   ->  KCl + MnCl2 + 4H2O + Cl2

Now there are eight hydrogen atoms on the right side of the equation, we must write 8 HCl.

KMnO4 + 8HCl    -> KCl + MnCl2 + 4H2O + Cl2

Of the eight chlorine atoms on the left, one is disposed of in KCl and two in MnCl2 leaving five free chlorine atoms. Therefore, the above equation becomes

KMnO₄+8HCl    -> KCl+MnCl₂+4H₂O+5/2 Cl₂

Equations  are  written  with  whole  number  coefficient  and  so  the  equation is multiplied through out by 2 to become

2KMnO₄+16 HCl ->  2KCl+2 MnCl₂+8H₂O+5Cl₂

Redox reactions [Reduction - oxidation]

In our daily life we come across process like fading of the colour of the clothes, burning of the combustible substances such as cooking gas, wood, coal, rusting of iron articles, etc. All such processes fall in the category of specific type of chemical reactions called reduction - oxidation (or) redox reactions. A large number of industrial processes like, electroplating, extraction of metals like aluminium and sodium, manufactures of caustic soda, etc., are also based upon the redox reactions. Redox reactions also form the basis of electrochemical and electrolytic cells. According to the classical concept, oxidation and reduction may be defined as,

Oxidation is a process of addition of oxygen (or) removal of hydrogen

Reduction is a process of removal of oxygen (or) addition of hydrogen.

Example

Reaction of Cl₂ and H₂S

H2S  +  Cl2  - >  2HCl  +  S

Oxidation : H2S   -> S

Reduction : Cl2 -> 2HCl 

In the above reaction, hydrogen is being removed from hydrogen sulphide (H₂S) and is being added to chlorine (Cl₂). Thus, H₂S is oxidised and Cl₂ is reduced.

Electronic concept of oxidation and Reduction

According to electronic concept, oxidation is a process in which an atom taking part in chemical reaction loses one or more electrons. The loss of electrons results in the increase of positive charge (or) decrease of negative of the species. For example.

Fe²⁺   -> Fe³⁺ + e^- [Increase of positive charge]

Cu  ->  Cu²⁺ + 2e^- [Increse of positive charge]

The species which undergo the loss of electrons during the reactions are called reducing agents or reductants. Fe²⁺ and Cu are reducing agents in the above example.

Reduction

Reduction is a process in which an atom (or) a group of atoms taking part in chemical reaction gains one (or) more electrons. The gain of electrons result in the decrease of positive charge (or) increase of negative charge of the species. For example,

Fe³⁺ + e^-  -> Fe²⁺ [Decrease of positive charges]

Zn²⁺ + 2e^-   ->  Zn [Decrease of positive charges]

The species which undergo gain of electrons during the reactions are called oxidising agents (or) oxidants. In the above reaction, Fe³⁺ and Zn²⁺ are oxidising agents.

Oxidation Number (or) Oxidation State

Oxidation number of the element is defined as the residual charge which its atom has (or) appears to have when all other atoms from the molecule are removed as ions.

Atoms can have positive, zero or negative values of oxidation numbers depending upon their state of combination.

General Rules for assigning Oxidation Number to an atom

The following rules are employed for determining oxidation number of the atoms.

1.    The oxidation number of the element in the free (or) elementary state is always Zero.

Oxidation number of Helium in  He = 0

Oxidation number of chlorine in Cl2 = 0

2.    The oxidation number of the element in monoatomic ion is equal to the charge on the ion.

3.   The oxidaton number of fluorine is always - 1 in all its compounds.

4.    Hydrogen is assigned oxidation number +1 in all its compounds except in metal hydrides. In metal hydrides like NaH, MgH2, CaH2, LiH, etc., the oxidation number of hydrogen is -1.

5.    Oxygen is assigned oxidation number -2 in most of its compounds, however in peroxides like H₂O₂, BaO₂, Na₂O₂, etc its oxidation number is -

6.     Similarly the exception also occurs in compounds of Fluorine and oxygen like OF₂ and O₂F₂ in which the oxidation number of oxygen is +2 and +1 respectively.

7.    The oxidation numbers of all the atoms in neutral molecule is Zero. In case of polyatomic ion the sum of oxidation numbers of all its atoms is equal to the charge on the ion.

8.    In binary compounds of metal and non-metal the metal atom has positive oxidation number while the non-metal atom has negative oxidation number. Example. Oxidation number of K in KI is +1 but oxidation number of I is - I.

9.    In binary compounds of non-metals, the more electronegative atom has negative oxidation number, but less electronegative atom has positive oxidation number. Example : Oxidation number of Cl in ClF3 is positive (+3) while that in ICl is negative (-1).

Problem

Calculate the oxidation number of underlined elements in the following species.

CO₂, Cr₂O₇^2-, Pb₃O₄, PO₄^3-

Solution

1. C in CO_2. Let oxidation number of C be x. Oxidation number of each O atom = -2. Sum of oxidation number of all atoms = x+2 (-2) Þ x - 4.

As it is neutral molecule, the sum must be equal to zero.

\ x-4 = 0 (or) x = + 4

2. Cr in Cr₂O₇^2-. Let oxidation number of Cr = x. Oxidation number of each oxygen atom =-2. Sum of oxidation number of all atoms

2x + 7(-2)  =  2x - 14

Sum of oxidation number must be equal to the charge on the ion.

Thus, 2x - 14  =  -2

2x  =  +12

x  =  12/2

x  =  6

Oxidation and Reduction in Terms of Oxidation Number

Oxidation                                                                     

"A  chemical  process  in  which  oxidation  number  of  the  element increases".                                                                            

Reduction                                                                     

"A  chemical  process  in  which  oxidation  number  of  the  element decreases".                                                                           

Eg.Reaction between H2S and Br2 to give HBr and Sulphur.

H2S (+1 -2) +   Br2 (0)  ->   2HBr  (+1 -1)  +  S  (0)

Decrease of Oxidation Number (Br) : Br2(0) -> 2HBr  (+1 -1)

Increase of oxidation Number (S)  : H2S (+1 -2) -> S  (0)

In the above reaction, the oxidation number of bromine decreases from 0 to -1, thus it is reduced. The oxidation number of S increases from -2 to 0. Hence H₂S is oxidised.

Under the concept of oxidation number, oxidising and reducing agent can be defined as follows.

i. Oxidising agent is a substance which undergoes decrease in the oxidation number of one of its elements.

ii. Reducing agent is a substance which undergoes increase in the oxidation number of one of its elements.

In the above reaction H₂S is reducing agent while Br₂ is oxidising agent.

Balancing Redox reaction by Oxidation Number Method

1.     The various steps involved in the balancing of redox equations according to this method are :

2.     Write skeleton equation and indicate oxidation number of each element and thus identify the elements undergoing change in oxidation number.

3.     Determine the increase and decrease of oxidation number per atom. Multiply the increase (or) decrease of oxidation number of atoms undergoing the change.

4.     Equalise the increase in oxidation number and decrease in oxidation number on the reactant side by multiplying the respective formulae with suitable integers.

5.     Balance the equation with respect to all atoms other than O and H atoms.

6.     Balance oxygen by adding equal number of water molecules to the side falling short of oxygen atoms.

7.     H atoms are balanced depending upon the medium in same way as followed in ion electron method.

Let us balance the following equations by oxidation number method.

MnO₂ + Cl^- ->  Mn²⁺ + Cl₂ +  H₂O (in acidic medium)

Step 1

MnO2 + Cl^-  - >  Mn²⁺ + Cl2 + H2O

Step 2

MnO2 (4+) + Cl (-1) -> Mn (+2) + Cl2 +    H2O

O.N. Decreases by 2 per Mn : MnO2 -> Mn

O.N. increases by 1 per Cl : Cl -> Cl2

Step 3

Equalise the increase / decrease in O.N by multiply MnO₂ by 1 and Cl^-1 by 2.

MnO₂ + 2 Cl^-   ->  Mn²⁺ + Cl₂ + H₂O

Step 4

Balance other atoms except H and O. Here they are all balanced.

Step 5

Balance O atoms by adding H₂O molecules to the side falling short of oxygen atoms.

MnO₂ + 2Cl^-    ->   Mn²⁺ + Cl₂ + H₂O + H₂O

Step 6

Balance H atoms by adding H⁺ ions to the side falling short of H atoms

MnO₂ + 2Cl^- + 4H⁺   ->   Mn²⁺ + Cl₂ + 2H₂O

Problems for practice

Balance the following equations

1.     Mg + NO₃^-    ->  Mg²⁺ + N₂O + H₂O (in acidic medium)

2.     Cr³⁺ + Na₂O₂   ->   CrO₄^-  + Na⁺

3.     S^2- + NO₃^-   ->    NO + S

4.     FeS + O2    ->   Fe2O3 + SO2 (molecular form)

5.     Cl₂ + OH^-   -> Cl^- + ClO₃^-  + H₂O