Chemistry CHEMICAL EQUATION AND STOICHIOMETRY

Stoichiometry :

The word 'Stoichiometry' is derived from two Greek words - Stoicheion (meaning element) and metron (meaning measure). Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction. Before understanding how to calculate the amounts of reactants required or those produced in a chemical reaction,let us study what information is available from the balanced chemical equation of a given reaction. Let us consider the combustion of methane. A balanced equation for this reaction is as given below:

`CH_4(g)+2O_2(g)-> CO_2(g)+2H_2O(g)`

The above balance reaction gives the following information:

For every `1` mole of `CH_4,` `2` mole of `O_2` will be required to produce `1` mole of `CO_2` and `2` moles of `H_2O`. this signifies `text(Mole - Mole relation)`

For every `16` gms of `CH_4` , `64` gms of `O_2` will be required to produce `44` gms of `CO_2` and `36` gms of `H_2O` this signifies `text(Mass-Mass relation)`

Ratio of moles of `CO_2 : H_2O` at any time `= 1 :2`

There will be no change in total mass of all reactants and products at any time for any chemical reaction.

Interpretation of Balanced Chemical Equations:

Once we get a balanced chemical equation then we can interpret a chemical equation by following ways:

a.) Mass - mass analysis

b.) Mass - volume analysis

c.) Volume - volume analysis

Mass- Mass Analysis:

Consider the reaction, `2KClO_3 -> 2KCl + 3O_2`

According to stoichiometry of the reaction
Mass - mass ratio: `2 xx 122.5 : 2 xx 74.5 : 3 xx32`

Or `(text(Mass of) (KClO_3))/(text(Mass of KCl)) = (2xx 122.5)/(2 xx 74.5)`

`(text(Mass of)( KClO_3))/(text(Mass of) (O_2)) = (2xx 122.5)/(3xx 32)`

Mass - Volume Analysis :

Now again consider decomposition of `KClO_3`

`2KClO_3 -> 2KCl + 3O_2`

Mass Volume ratio, `2 xx 122.5 gm : 2 xx 74.5 gm: 3 xx 22.4L` at S.T.P.

We can use two relation for volume of oxygen.

`(text(Mass of) quad KClO_3)/(text(volume of) quadO_2 quad at quad S.T.P.) = (2 xx 122.5)/(3 xx 22.4L)`

And `(text(Mass of)quad KCl)/(text(volume of) quadO_2 quad at quad S.T.P.) = (2 xx74.5)/(3 xx 22.4L)`

Volume-Volume Relationship:

It relates the volume of gaseous species (reactants or product) with the volume of another gaseous species (reactant or product) involved in a chemical reaction.

Limiting Reagent:

In many situations one of the reactants is present in excess therefore some of this reactant is left over on completion of the reaction. For example, consider the combustion of hydrogen.

`2H_2(g) + O_2(g) -> 2H_2O(g)`

Suppose that `2` moles of `H_2` and `2` moles of `O_2` are available for reaction. It follows from the equation that only `1` mole of `O_2` is required for complete combustion of `2` moles of `H_2` ; `1` mole of `O_2` will, therefore, be left over on completion of the reaction. The amount of the product obtained is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is completely consumed during the reaction.

In the above example `H_2` is the limiting reagent. The amount of `H_2O` formed will, therefore, be determined by the amount of `H_2`. Since `2` moles of `H_2` are taken, it will form `2` moles of `H_2O` on combustion.

Calculation of Limiting Reagent :

Method-I : By calculating the required amount by the equation and comparing it with given amount.
[Useful when only two reactant are there)

Method-II : By calculating amount of any one product obtained taking each reactant one by one irrespective of other reactants. The one giving least product is limiting reagent.

Method-III : Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. [Useful when number of reactants are more than two.)

Percentage Yield :

In general, when a reaction is carried out in the laboratory we do not obtain the theoretical amount of product. The amount of product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield, the % yield can be calculated by the following formula-

`%` Yield = `(text(actual yield)/text(theoritical yield)) xx100`

Principle of Atom Conservation (POAC):

In chemical reaction atoms are conserved, so moles of atoms shall also be conserved. This is known as principle of atomic
conservation. This principle is helpful in solving problems of nearly all stoichiometric calculations e.g.

`KClO_3(s) -> KCl(s) + O_2(g)`

Applying POAC for `K` atoms,

Moles of `K` atoms in `KClO_3` = Moles of `K` atoms in `KCl`

Since one mole of `KClO_3` contains `1` mol of `K` atom. Similarly, `1` mol of `KCl` contains one mole of `K` atoms.

`n_(KClO_3) = n_(KCl)` `=>` `w_(KClO_3)/M_(KClO_3) = w_(KCl)/M_(KCl)` (Mass-mass relationship)

Applying POAC for `O` atoms Moles of `O` atom in `KClO_3` = Moles of `O` atom in `O_2`

`3xx n_(KClO_3) = 2xx n_(O_2)`

`=> 3xx w_(KClO_3)/M_(KClO_3)` = `2xx ((vol. quad of quad O_2 quad at quad STP)/ text(Standard Molar volume))`

(Mass volume relationship of reactant and product)

In this way applying POAC we can break the chemical equation into a number of arithmetic equations without balancing the chemical equation. Moreover number of reactions and their sequence from reactants to products are not required. It is important to note that POAC can be applied for the atoms which remain conserved in chemical reaction.

Please note that the balanced reaction is essential in Mole method of solving. Only while using POAC (which would be applicable only in certain cases), balanced reaction is not required.

You can use POAC for all atoms in the compound only if all the reactants and products are known.