How many moles of $H_2{SO}_4$ can be reduced to ${SO}_2$ by 2 moles of Aluminium?

 

${\mathrm{H}}_2{\mathrm{SO}}_4+2{\mathrm{e}}^{-}⟶{\mathrm{SO}}_2+{\mathrm{H}}_2\mathrm{O}$

1 mole of $H_2{SO}_4$ reduced to ${SO}_2$ , 2e^- are required

$Al⟶A{l}^{+3}+3e^{-}$

Moles of $H_2{SO}_{4}required=\frac{6e}{2e}=3$

Therefore, moles of $H_2{SO}_4$ is 3.