# How to calculate n factor for salts in this reaction?

My source says : For salts which react such that no atom undergoes change in oxidation state , n-factor for such salt is the total moles of cationic /anionic species replaced in 1 mole of the salt .

For Example $$\ce{Na3PO4 + BaCl2 -> NaCl + Ba3(PO4)3}$$

N factor of $$\ce { Na3PO4}$$ is 3.

Please can anybody give more examples and explain that to me ?

$$\ce{2 Na3PO4 + 3 BaCl2 -> 6 NaCl + Ba3(PO4)2}$$
Now this is clearly evident that dividing whole equation by 2 we get 1 mole of $$\ce{Na3PO4}$$ and at the right hand side 3 moles of $$\ce{NaCl}.$$ Therefore, 1 mole of phosphate ion is replaced by 3 moles of $$\ce{Cl-}$$ ion.
Hence $$n\text{-factor} = 3.$$
In reactions like this where there's no change in oxidation number, the $$n\text{-factor}$$ will always be the net positive charge in 1 molecule of that compound, in this case since there are 3 $$\ce{Na}$$ ions in sodium phosphate, the $$n\text{-factor} = 3 × 1 = 3.$$