In the following reaction, equal moles of ${\mathrm{Ca}}_3{\mathrm{P}}_2$ & ${\mathrm{H}}_2\mathrm{O}$ is taken then the limiting reagent is  ${\mathrm{Ca}}_3{\mathrm{P}}_2+6{\mathrm{H}}_2\mathrm{O}\to 3\mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2{\mathrm{PH}}_3$

${\mathrm{Ca}}_3{\mathrm{P}}_2+6{\mathrm{H}}_2\mathrm{O}\to 3\mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2{\mathrm{PH}}_3$

1 mole ${\mathrm{Ca}}_3{\mathrm{P}}_2\to 3 \mathrm{moles} \to 2 \mathrm{moles}$

${\mathrm{Ca}}_3{\mathrm{P}}_2=3 \mathrm{moles} =2 \mathrm{moles}$

6 mole ${\mathrm{H}}_2\mathrm{O}\to 3 \mathrm{moles}\to 2 \mathrm{moles}$

1 mole ${\mathrm{H}}_2\mathrm{O}=0.5 \mathrm{moles}=0.33 \mathrm{moles}$

$\therefore {\mathrm{H}}_2\mathrm{O}$ is limiting reagent. Since it gives less amount of product.