0.5 mole of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is mixed with 0.2 mole of $\mathrm{Ca}(\mathrm{OH}{)}_2$. The maximum number of moles of ${\mathrm{CaSO}}_4$ formed is:

${\mathrm{H}}_2{\mathrm{SO}}_4+\mathrm{Ca}(\mathrm{OH}{)}_2\to {\mathrm{CaSO}}_4+2{\mathrm{H}}_2\mathrm{O}$

$\begin{array}{l}0.5 \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.2\end{array} - -$

                  (LR)                    0.2

We have 0.5 moles of ${\mathrm{H}}_2$${\mathrm{SO}}_4$

and 0.2 moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$

So, Calcium hydroxide is limiting here and 0.2 moles of ${\mathrm{CaSO}}_4$ will be produced.