For the following reactions: 

I.  ${\left(N{H}_4\right)}_{2}S{O}_4+2NaOH\stackrel{40\%}{\to }N{a}_{2}S{O}_3+2H_{2}O+2N{H}_3$

II. $N{H}_3+HCl\stackrel{80\%}{\to }N{H}_{4}Cl$

If $4g$ of $NaOH$ is taken then 

Produced moles of $N{H}_{4}Cl$ (in reaction $I{I}^{nd}$ reaction) are 1.6 times of produced moles of $N{a}_{2}S{O}_4$ (in reaction I).  

Reacting moles of $HCl$ (in reaction II) is lesser than reacting moles $NaOH$ (in reaction I).

Reacting moles of $HCl$ (in reaction II) is 20% lesser than original ${\left(N{H}_4\right)}_{2}S{O}_4$ moles.

Produced mass of $N{H}_{4}Cl$ is $2.71g.$