\(\overset { (+2)(-2) }{ { CuO }_{ (s) } } +\overset { (0) }{ { H }_{ { 2 }_{ (g) } } } \rightarrow \overset { (0) }{ { Cu }_{ (s) } } +\overset { (+1)(+2) }{ { H }_{ 2 }O_{ (g) } } .\)
In the above reaction, oxygen is removed from CuO. So CuO gets reduced. Oxygen is added to H₂ to form water. So H₂ gets oxidised. i.e. In CuO, oxidation number of Cu +2 is reduced to 0 whereas in H₂, oxidation of H₂ O is increased to + 1. So the above reaction is a redox reaction.

Molar mass of ¹²C =  12.00 g mol^-1.
Mass of 6.023\(\times\)10²³ carbon atom =  12.0 g
\(\therefore\) Mass of 1 carbon atom \(={12\over6.023\times{10}^{23}}\)
= 1.992\(\times\)10^-23 g.

\(\underset { 2\ volumes }{ { 2H }_{ { 2 }_{ (g) } } } +\underset { 1\ volumes }{ { O }_{ { 2 }_{ (g) } } } \rightarrow \underset { 2\ volumes }{ { 2H }_{ 2 }{ O }_{ (g) } } \)
Thus 2 volumes of H₂ reacts with 1 volume of O₂ to produce 2 volumes of H₂O (g)
\(\therefore\) 10 volumes of H₂ would react with 5 volumes of O₂ to produce 10 volumes of H₂O(g).
Thus 10 volumes of H₂O will be produced.

2x + 7 (-2) = -2
2x - 14 = -2
2x = +12
\(\therefore\)  x = + 6
Oxidation state of Cr = +6.

1 (+1) + x + 4 (-2) = 0
x -7 = 0
\(\therefore\)  x = + 7
Oxidation state of Mn = +7.

(a) Burning of cooking gas, wood
(b) Rusting of iron articles
(c) Electroplating
(d) Galvanic and electrolytic cells

SF₆ is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of S is allowed to react with 12 moles of Fluorine.
\({ S }_{ (1) }+{ 3F }_{ 2_{ (g) } }\rightarrow { SF }_{ { 6 }_{ (g) } }\)
As per the stoichiometric reaction, one mole of S reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.
\(\therefore\)  It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.

A large excess of one reactant is supplied to ensure the more expensive reactant is completely converted to the desired product. The reactant used up first in a reaction is called the limiting reagent.

\(\underset { (1\times 22.4L) }{ { H }_{ { 2 }_{ (g) } } } +\underset { (1\times 22.4L) }{ Cl_{ { 2 }_{ (g) } } } \rightarrow \underset { (2\times 22.4L) }{ 2HCl_{ _{ (g) } } } \) 
2 x 22.4 L of HCI is produced by 22.4 L of Cl₂.
\(\therefore\) 89.6 L of HCl will be produced by

= 44.8 L of chlorine.

CaCO_3(s) \(\rightarrow\) CaO_(s) + CO_(g)
100g                       22.4L
100 g of CaCO₃ produces 22.4 L of CO₂.
\(\therefore\) 25 g of CaCO₃ will purchase  = \(\frac { 22.4 }{ 100 } \times 25\)
= 5.6 L of CO₂.

3H₂ + N₂ \(\rightarrow\) 2NH₃
A per stoichiometric equation,
No. of moles of hydrogen required for 2 moles of ammonia = 3 moles
No. of moles of hydrogen required for 20 moles of ammonia = \(\frac { 3 }{ 2 } \times 20=30\) moles.

Sulphuric acid = H₂SO₄
Molar mass of sulphuric acid = 2 + 32 + 64 = 96
Basicity of sulphuric acid  = 2
equivalent mass of acid = \(\frac { Molar\ mass\ of\ an\ acid }{ Basicity } \)
= \(\frac { 96 }{ 2 } \) = 49g eq^-1.

Phosphate ion (PO₄^3-).
Equivalent mass of Phosphate ion = \(\frac { Molar\ mass\ of\ sulphate\ ion }{ charge } \)
= \(\frac { 31+64 }{ 2 } =\frac { 95 }{ 3 } =31.6\)
= 31.6g eq^-1.

Sulphate ion (SO₄^2-).
Equivalent mass of Sulphate ion = \(\frac { Molar\ mass\ of\ sulphate\ ion }{ charge } \)
\(=\frac { 32+64 }{ 2 } =\frac { 96 }{ 2 } \)
= 48 g eq^-1.

Equivalent mass = \(\frac { Atomic\ mass }{ Velency } \)
Equivalent mass of Copper = \(\frac { 63.5 }{ 2 } =31.75\ g\)eq^-1.

No.of moles = \(\frac { mass\ of\ the\ substance }{ Molar\ mass\ of\ the\ substanc } =\frac { W }{ M } \)
Molar mass of ethane (C₂H₆) = 24 + 6 = 30
\(\therefore\) Number of moles in 60 g of ethane = \(\frac { 60 }{ 30 } \) = 2 moles.

Element	No.of atoms	Relative atomic mass	Total relative atomic mass
H	2	1	2
S	1	32	32
O	4	16	64
Molecular mass of sulphuric acid = 98

Molecular mass of a substance (element or compound) represents the number of times the molecule of that substance is heavier than 1/12^th of the mass of an atom of C-12 isotope. Molecular mass = 2 x Vapour density

Chlorine molecule has two isotopes as in ₁₇Cl³⁵, ₁₇Cl³⁷in the ratio of 77 : 23, so when we are calculating the average atomic mass, it becomes fractional.
The average relative atomic mass of Chlorine = \(\frac { \left( 35\times 77 \right)+ \left( 37\times 23 \right) }{ 100 } \)
= 35.46 amu.

	Molecule	Compound
i)	A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence.	A molecule which contains two or more atoms of different elements are called a compound molecule
ii)	e.g. Na - Monoatomic molecule  O2 - Diatomic molecule P4 - Polyatomic molecule	e.g. CO2 - Carbon dioxid CH4 - Methane H2O- Water

An atom is the ultimate smallest electrically neutral, being made up of fundamental particles such as proton, neutron and electron.
An element consists of only one type of atoms. Elements are further divided into metals, non-metals, and metalloids

Gaseous state of matter at very high temperature containing gaseous ions and free electron is referred to as the Plasma state. e.g. Lightning.

States of matter are inter convertible by changing temperature and pressure.
\(Solid\overset { Heat }{ \underset { Cool }{ \rightleftharpoons } } Liquid\overset { Heat }{ \underset { Cool }{ \rightleftharpoons } } Gas.\)

A matter is anything which has mass and occupies space. Matters exist in all three states such as solid, liquid and gas.

Empirical Formula	Molecular Formula
It is the simplest formula	It is the actual formula
It shows the ratio of number of atoms of different elements in one molecule of the compound.	It shows the actual number of different types of atoms present in one molecule of the compound

Hydrated sodium carbonate = Na₂CO₃.10H₂O
Molecular mass of Na₂CO₃ = (23 x 2) + (12 x 1) + (16 x 13) + (1 x 20)
= 46 +12 + 208 + 20
= 286
Equivelent mass of Na₂CO₃.10H₂O = \(\frac { Molecular\ mass }{ acidity } \)
= \(\frac { 286 }{ 2 } =143\)

Acidity : The number of hydroxyl ions present in one mole of a base is known as the acidity of the base.
Basicity : The number of replaceable hydrogen atoms present in a molecule of the acid is referred to as its basicity.

Molar volume is the volume occupied by one mole of a substance in the gaseous state at STP. It is equal to 2.24 x 1O^-2m³(22.4 L).

= (12 x 12) + (22 x 1) + (11 x 16)
= 342 g /mol

= (39 x 2) + (2 x 52) + (7 x 16)
= 294 g mol.

= (1 x 39) + (1 x 55) + (4 x 16)
= 158 g / mol

(i) C₂H₅OH : (2 x 12) + (5 x 1) + (1 x 16) + (1 x 1) = 46 g
(ii) KMnO₄ : (1 x 39) + (1 x 55) + (4 x 16) = 158 g
(iii) K₂Cr₂O₇ : (2 x 39) + (2 x 52) + (7 x 16) = 294 g
(iv) C₁₂H₂₂O₁₁ : (12 x 12) + (22 x 1) + (11 x 16) = 342 g

Molecular mass = Density x Molar volume
Molar volume of CO₂ = 2.24 x 10^-2m³
Density of CO₂ 1.977 kg m^-3
\(\therefore \text { Molecular mass of } \mathrm{CO}_{2}=1.977 \times 10^{3} \mathrm{~g} \mathrm{~m}^{\not -3} \times 2.24 \times 10^{-2} \mathrm{~m}^{\not -3}\)
= 1.977 X 10¹ x 2.24
= 44g

\(Equivalent \ mass ={sum \ of \ the \ atomic \ masses \ of \ atoms \ present \ in \ the \ ion \over charge \ on \ the \ ion}\)
For CO₃ ^-2 ion
\(Equivalent \ mass ={1 x atomic \ mass \ of C + 3 x atomic \ mass \ of 'O'\over 2}\)
\(={12+3\times 16\over 2}={60\over2}=30\ g\ eq^{-1}\)

equivalent mass = \(Molar \ mass \over Basically\)
For H₃PO₄ equiavalent mass = \(3 \times atomic \ mass \ of \ 'H' + 1 \times atomic \ mass \ of \ phosphorous \ + 4 \times atomic \ mass \ of 'O '\over3\)
\(=3\times 1+ 1\times 31+4\times16\over3\)
\(={3+31+64\over3}={98\over3}=32.66\ g\ of\ eq^{-1}\)

Molar mass of K₂Cr₂O₇ = 2 x 39 + 2 x 52 + 7 x 16
= 78 + 104 + 112 = 294
In acid medium Cr₂O₇^-2 is reduced to Cr⁺³
Cr₂O₇ ^-2 + 6e + 14H + \(\rightarrow\) 2cr⁺³ + 7H₂O
equivalent mass of K₂Cr₂O₇ = \({Molar \ mass \over No, \ of \ electrons \ taken \ Up }\)
\(={294\over6}=49\)
\(\therefore\) 'n' factor \(={294\over49}=6 eq.mol^{-1}\)

'n' factor for Ba(OH)₂ = \({Molar \ mass \ of \ Ba(OH)2 \over equivalent \ mass \ of Ba(OH)2 }\)
Molar mass of Ba(OH)₂ = 137 + 2 x 16 + 2 x 1
= 137 + 32 + 2 = 171
Equivalent mass of Ba(OH)₂ = \({molar \ mass \over acidity}\)
\(={171\over2}=85.5\)
\(\therefore\)'n' factor for Ba(OH)₂ \(={171\over85.5}=2\)

'n' factor for H₂SO₄ =\({Molar \ mass \ of \ the \ acid \over equivalent \ mass \ of \ the \ acid}\)
\({98\over49}=2\)

\(Equivalence \ factor (n) ={Molar \ mass \ (g mol^{-I})\over equivalent \ mass \ of \ the \ acid}\)
The usefulness of this factor is that the equivalent masses of all the substances can be calculated whether it is an acid, base, salt, or an oxidising or reducing agent.

Gram equivalent mass of an element, compound or ion is the mass of their species that combines or displaces 1.008g hydrogen or 8g oxy,gen or 35.5g chlorine.

Stoichiometry gives the numerical relationship between chemical quantities in a balanced chemical equation.

2 mole of H₂ react with 1mole of O₂.
1.5 mole of H react with 0.75 mole of O₂.
Excess O₂ = 0.9062 - 0.75 = 0.1562 mol.

2 mole of H₂ give 2 mole of H₂O.
1.5 mole of H₂ will give 1.5 mol of H₂O.

Molecular formula : Molecular formula of a compound gives the actual number of different atoms present in one molecule

Empirical formula : It is the simplest formula of a compound which gives the ratio of the number of different atoms present in one molecule of the compound.

\(\underset { 2\quad mol }{ { 2H }_{ 2 } } +\underset { 1\quad mol }{ { { O }_{ 2 } } } \rightarrow \underset { 2\quad mol }{ { 2H }_{ 2 }O } \)
Number of mole of H₂ taken = \(\frac{3}{2}\) = 1.5
Number of mole of O₂ taken = \(\frac{29}{32}\) = 0.9062
\(\therefore\) 2 mole of H₂ react with 1 mole of O₂
\(\therefore\) 1.5 mol of H₂ will react with = \(\frac{1}{2}\times \) 1.5 = 0.75 mole of O₂
Available mole of O₂ (0.9062) > required mol of O₂ (0.75)
Thus, O₂ is in excess and therefore, H₂ will be the limiting reagent.

When a reaction is carried out using non-stoichiometric quantities of the reactants, the product yield will be determined by the reactant that is completely consumed. This reagent is called as the limiting reagent.

Molar Mass of K₂SO₄ = \( \begin{cases} 2x\ atomic\ \ mass\ of\ K+1\times atomic\ \ mass\ of\ 's' \\ +\ 4x\ atomic\ \ mass\ of\ 'O' \end{cases}\)
= 2\(\times\)40 +1\(\times\)32 + 4\(\times\)16 = 80 + 32 + 64 = 176
Equivalent Mass of salt = \(\frac { atomic\ \ mass\ \ of\ \ cation }{ Velency } +\frac { sum\ \ of\ \ atomic\ \ equivelent\ \ mass\ \ of\ \ anion }{ valency } \)
= \(\frac { 40 }{ 1 } +\frac { 32+64 }{ 2 } \)
= 40 + 48 = 88g eq^-1 

Molar Mass of NaOH = 23 + 16 + 1 = 40
Acidity = 1
Equivalent Mass = \(\frac { Molar\ mass }{ acidity } =\frac { 40 }{ 1 } =40geq^{ -1 }\)

Molar Mass of Al(OH)₃ = \(\begin{cases} atomic\ \ mass\ 'al'+3x\ atomic\ \ mass\ \ of \ 'O' \\ +3x\ atomic\ \ mass\ \ of\ \ 'h' \end{cases}\)
= 27 + 3\(\times\)16 + 3\(\times\)1 = 78
Acidity = 3
Equivalent mass of AL(OH)₃ = \(\frac { Molar\quad mass }{ acidity } =\frac { 78 }{ 3 } =26geq^{ -1 }\)