How to calculate n factor for O2 and H2 in this reaction?
Here is the unbalanced reaction
$H_2 +O_2----&gt; H_20$
I hope you know how to calculate oxidation number
 On product side
In $H_20$ oxidation number of $H$ is  $+1$ and $O$ is $-2$
On reactant side
oxidation number of $H$ and $O$ is  $0$
So n-factor is magnitude in change of oxidation number
So n factor of $H$ is $1-0$ i.e $1$ and $O$ is $-2-0$ i.e $2$(magnitude)
Now to balance the reaction (for 2 reactant rex) you just interchange the n factors
You get 
$1O_2 +2H_2 -----&gt; 2H_20$
Now to calculate equivalent weight formula is 
$E.w=molar mass/N.F$