1 mole of each of  $Ca{\left(OH\right)}_2$ and $H_{3}P{O}_4$ are allowed to react under dilute conditions.  The maximum number of moles of $C{a}_3{\left(P{O}_4\right)}_2$ formed is

$3Ca{\left(OH\right)}_2+2H_{3}P{O}_4\to C{a}_3{\left(P{O}_4\right)}_2+H_{2}O)$

3 mole          2 moles

1 mole          1 mole

$\therefore Ca{\left(OH\right)}_2$ is limiting reagent

Number of moles of  $C{a}_3{\left(P{O}_4\right)}_2$  formed $=\frac{1}{3}$  moles.