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How to calculate n factor in non-redox reactions and other such reactions?

Answer
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Hint: We need to remember that the valency factor or conversion factor is the n factor. The calculation of valency factor depends upon the type of reactant which we are using to find the n factor. For non-redox reactions, n factor of a substance is equal to the product of displaced mole and charge of the product.

Complete step by step answer:
We need to know that the equivalent weight of the substance is calculated by molecular weightn.
This is often inferred that the no. of gram equivalents of a substance is equal to n×number of moles and also Normality =n×molality.
It is appropriate to use the law of gram equivalence to solve problems supporting chemical reactions. Consistent with this law, the number of gram equivalents of all reactants are adequate to the amount of gram equivalents of all products, assuming that each one the reactants are undergoing within the reaction.
We can use the above law conveniently to unravel problems without recurring to understand much about the reactions. For this we’d like to possess a good understanding of the n factor of the substance.
Non-redox reaction: It is a chemical reaction in which neither oxidation nor reduction takes place.
For Acids: Acids are the species which provides H+ ions when dissolved during the solvent.
For acids, the number of H+ ions replaced by one mole of acid during a reaction is called n factor. It is important to note that the n factor for acid isn’t adequate for its basicity, i.e. the amount of moles of replaceable H+ ions present in one mole or acid.
For example,
N factor of HCl = 1
N factor of HNO3 = 1
N factor of H2SO4 = 1 or 2, depending on the extent of reaction it undergoes.
H2SO4+NaOHNaHSO4+H2O
In the above reaction, although one mole of H2SO4 has two replaceable hydrogen atoms but during this reaction H2SO4 has given just one H+ ion, so n factor for this would be 1.
H2SO4+2NaOHNa2SO4+2H2O
The n factor for H2SO4 during the above reaction would be 2.
Similarly,
H2SO3 n factor = 1 or 2
H2CO3 n factor = 1 or 2
H2PO4 n factor = 1 or 2 or 3
H3PO3 n factor = 1 or 2, it is because one among the hydrogen isn’t replaceable in H3PO3. This will be seen using structure.
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H3BO3 n factor = 1.
The three hydrogen atoms are replaceable in H3BO3 . But boron is an electron deficient atom, so it acts as a Lewis acid. When H3BO3 is reacted with water, H2O, the oxygen atom of water attacks the boron atom, though its lone pair. The equation is follows,
H3BO3+2H2O[B(OH)4]+H3O+
Thus, one mole of H3BO3 in the solution gives just one mole of H+, so its n factor is one.
Bases: Bases are the species that provides OH ions when dissolved during a solvent.
For bases, the number of OH ions replaced by one mole of base during a reaction is called n factor.
It is important to note that n factor isn’t adequate to its acidity, i.e. the amount of moles of replaceable OH ions present in one mole of a base.
For example,
NaOH n factor = 1
Al(OH)3 n factor = 1 or 2 or 3
NH4OH n factor = 1
Zn(OH)2 n factor = 1 or 2
Ca(OH)2 n factor = 1 or 2

Note:
 The n factor for salt is, a salt reacts such that no atom of the salt undergoes any change in oxidation number. For example,
2AgNO3+MgCl2Mg(NO3)2+2AgCl
In the above reaction, it is seen that the oxidation number of Ag,N,O,Mg and Cl remains the same even after the reaction even within the product.
The n factor for such a salt is that the total charge on cation or anion.