$4\mathrm{Zn}+10{\mathrm{HNO}}_3⟶$ $4\mathrm{Zn}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{NH}}_4{\mathrm{NO}}_3+3{\mathrm{H}}_2\mathrm{O}$In this reaction one mole of HNO₃ is reduced by

From eq.

10 moles of HNO₃ is reduced by 4moles of Zn

1mole of HNO₃ is reduced by     ?

                 =$\frac{4}{10}=0.4\mathrm{moles}$ of Zn = 0.4x65=26g