1 mole of ${\mathrm{H}}_2{\mathrm{SO}}_4$ will exactly neutralise :

Acid normality equals base normality. As a result, 2 moles of NaOH will precisely neutralise 1 mole of H₂SO₄.

$\begin{array}{l}2{\mathrm{NH}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\\ \mathrm{Ba}(\mathrm{OH}{)}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{BaSO}}_4+2{\mathrm{H}}_2\mathrm{O}\\ 2\mathrm{KOH}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}\end{array}$

 Thus, the correct answer is A, B, D.