Answer:
Two moles.
Explanation:
Sulphuric (sulfuric) acid [tex]\rm H_2SO_4[/tex] is a diprotic acid. When one mole of [tex]\rm H_2SO_4[/tex] molecules dissolve in water, two moles of [tex]\rm H^{+}[/tex] ions would be produced.
[tex]\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}[/tex].
On the other hand, sodium hydroxide [tex]\rm NaOH[/tex] is a monoprotic base. When one mole of [tex]\rm NaOH[/tex] formula units dissolve in water, only one mole of hydroxide ions [tex]\rm OH^{-}[/tex] would be produced.
[tex]\rm NaOH \to Na^{+} + OH^{-}[/tex].
Note that [tex]\rm H^{+}[/tex] and [tex]\rm OH^{-}[/tex] react at a one-to-one ratio:
[tex]\rm H^{+} + OH^{-} \to H_2O[/tex].
As a result, it would take [tex]2\; \rm mol[/tex] of [tex]\rm OH^{-}[/tex] to react with the [tex]\rm 2\; mol[/tex] of [tex]\rm H^{+}[/tex] that was released when [tex]1\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex] is dissolved in water. Since one mole of [tex]\rm NaOH[/tex] formula units could produce only one mole of [tex]\rm OH^{-}[/tex], it would take [tex]\rm 2\; mol[/tex] of [tex]\rm NaOH[/tex] formula units to produce that [tex]2\; \rm mol[/tex] of [tex]\rm OH^{-}[/tex] for reacting with [tex]1\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex].
