Chemistry Concepts - Mole concept, Eudiometry, POAC, Limiting Reagent, Chemical Equivalence and Disproportionation reaction
Mole concept:
The mole is defined as the amount of a substance containing as many atoms, molecules, ions, electrons or other elementry entities as there are carbon atoms in exactly 12 g of 12C.
NA = 6.022 X 1023
The following are the definitions of ‘mole’ represented in the form of equations:
(1) no. of moles of molecules/atoms = weight in g
molecular/atomic wt
(2) no. of moles of gases = volume at NTP
standard molar volume (i.e. 22.4 L)
(3) no. of moles of entities = no. of entities
Avagadro constant (i.e. 6.022 X 1023)
(4) no. of moles of solute = molarity X volume of solution in L
no. of millimoles = molarity X volume of solution in mL
(5) For a compound MxNy, x moles of N = y moles of M
Note: 1 mole is a fixed no. of particles but not a fixed weight.
Principle of Atom Conservation (POAC):
This principle states that the moles of atoms of an element are conserved throughout the reaction.
This means, moles of an element in reactants = moles of the element in products.
eg. CO + O2 —–> CO2
In this moles of C in CO = moles of C in CO2
This means, moles of CO = moles of CO2.
Similarly, for O,
moles of O in CO + moles of O in O2 = moles of CO2
Thus, moles of CO + 2(moles of O2) = 2(moles of CO2)
Advantages of Mole Method over other Methods:
(1) Balancing of chemical equations is not required in the majority of problems as the method of balancing the chemical equation is based on the principle of conservation of atom conservation.
(2) Number of reactions and their sequences, leading from reactants to products, need not be given.
Note: Whenever balanced chemical equation is given, mole method is very useful.
eg. 2KClO3 —–> 2KCl + 3O2
Thus, 2(moles of KClO3) = 2(moles of KCl)
3(moles of KClO3) = 2(moles of O2)
3(moles of KCl) = 2(moles of O2)
Concept of Limiting Reagent:
In single-reactant reactions, the calculations are carried out with only that amount of the reactant which has converted to the product.
In the reactions where more than on reactant is involved, one has to first identify the limiting reactant, i.e., the reactant which is completely consumed. All calculations are to be carried out with the amount of the limiting reactant only.
Initially ….. 5 moles 12 moles 0 moles
A + 2B —–> 4C
If A is limiting reactant: moles of C produced = 20
If B is limiting reactant: moles of C produced = 24
The reactant producing the least number of moles of the product is the limiting reactant and hence A is the limiting reactant. Thus,
Initially ….. 5 moles 12 moles 0 moles
A + 2B ——> 4C
Finally ….. 0 moles 4 moles 20 moles
The limiting reactant can also be ascertained by knowing the initial number of equivalents of each reactant. The reactant with the least number of equivalents is the limiting reactant.
Note: Generally, when the initial moles / equivalents of all the reactants is given, then the question is based on the concept of limiting reagent.
Eudiometry:
Reactants Absorbed gases
KOH solution CO2 & SO2
Alkaline Pyrogallol O2
Soln of ammonical cuprous chloride CO
and if the mixture is cooled, H2O vapour changes to liquid.
Hence, When temperature is above 100 0C (373 K) then only volume of H2O is considered.
CxHy + (x + (y/4)) O2 —–> x CO2 + (y/2) H2O.
Eudiometry is mainly based on Avogadro’s law,
A(g) + B(g) —–> C(g) + D(g)
a volume b volume c volume d volume
a moles b moles c moles d moles
Chemical Equivalence:
One equivalent of any substance is the amount of that substance that combines with 1 g H or displaces 1 g H.
no. of equivalents = weight
equivalent weight
And, in a compound, MxNy, no of equivalents of M = no of equivalents of N
Equivalent weights of some elements :
H - 1g; O - 8g, Cl - 35.5g …..
For metals in hydrides, oxides and chlorids,
eq. wt = atomic weight
valency (in the compd)
Equivalent weight = Molecular weight
n
Hence, no. of equivalents = no. of moles X n
Hence, Normality = Molarity X n
Law of chemical equivalence:
aA + bB —–> cC + dD
eq of A = eq of B = eq of C = eq of D
Neutralisation reactions:
For acids, n = no of replacable H+ ions
Thus, for HCl - n = 1
H2SO4 - n = 2
H3PO4 - n = 3
H3PO3 - n = 2
Thus, for NaOH - n = 1
Ca(OH)2 - n = 2
For salts, n = total charge on cations or anions
Thus, for NaCl - n = 1
CaCl2 - n = 2
Ca3(PO4)2 - n = 6
Note: When the reactions are provided, then take the n - factor according to the reaction.
eg. H3PO4 + NaOH —–> NaHPO4 + H2O
Here only 1 H is replaced, n = 1
When 2 or more indicators are mentioned in the question, then the reaction takes place in two phases.
Phenolphthalein shows colour change after the first step and Methyl st1:cityst1:placeOrange/st1:place/st1:city shows the colour change after the completion of the complete reaction.
2Na2CO3 + H2SO4 Ã 2NaHCO3 + Na2SO4
n = 1 n = 2
2NaHCO3 + H2SO4 Ã Na2SO4 + 2H2CO3
n = 1 n = 2
Redox Reactions:
n = total change in oxidation number for the molecule / ion
total number of electrons gained / lost by the molecule
K2Cr2O7 (oxidizing agent) Cr +6 Ã Cr +3 n = 6
Na2S2O3 (reducing agent) S +2 Ã S +2.5 n = 1
H2C2O4 (reducing agent) Ã CO2 n = 2
NaHC2O4 (reducing agent) Ã CO2 n = 2
Na2C2O4 (reducing agent) Ã CO2 n = 2
Note: H2C2O4 NaHC2O4
For redox: n = 2 n = 2
For neutralistion n = 2 n = 1
KMnO4 (oxidizing agent)
In acidic medium, KMnO4 Ã Mn +2 n = 5
In basic/neutral medium, KMnO4 Ã MnO2 n = 3
In limited and controlled amount of base, KMnO4 Ã K2MnO4 n = 1
H2O2 (oxygen ? oxidation state = 1)
H2O2 Ã O2 (as a reducing agent) n = 2
H2O2 Ã H2O (as a oxidising agent) n = 2
Note: Oxidation number of s-block do not change.
For, FeC2O4, (in presence of oxidizing agent) n = 3 as
Fe +2 Ã Fe +3 + e-
2C +3 Ã 2C +4 + 2e-
Thus, total transfer of electrons = 3
Similarly, for H2C2O4.NaHC2O4.xH2O Ã CO2 n = 4
And for H2C2O4.FeC2O4.xH2O Ã CO2 n = 5
Disproportionation reaction:
Simultaneous oxidation & reduction of an element present in the molecule or ion in the same reaction.
In this,
Eq. wt. of molecule= (Eq. wt. due to oxidation) + (Eq. wt. due to reduction)
And ( 1 / noverall) = ( 1 / noxi ) + ( 1 / n red