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 $$\ce{Na3PO4 + BaCl2  -&gt; NaCl + Ba3(PO4)3}$$

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

Please can anybody give more examples and explain that to me ?
$$\ce{2 Na3PO4 + 3 BaCl2 -&gt; 6 NaCl + Ba3(PO4)2}$$
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 $n\text{-factor} = 3 × 1 = 3.$