Looking at the equation .look at the moles noted below
Ca(OH)_2=1mol
NH_2Cl=2mol.
CaCL_2=1mol
NH_2=2mol
H_20=2mol.
#a
There is 2 moles of NH2Cl.
Hence limiting reagent is Ca(OH)_2
#2
Moles at reactant=3
Moles at product=5
Moles left:-
[tex]\\ \sf\longmapsto 5-3=2mol[/tex]
#d
[tex]\\ \sf\longmapsto pV=nRT[/tex]
[tex]\\ \sf\longmapsto 1.5V=2(8.3)(27)[/tex]
[tex]\\ \sf\longmapsto 1.5V=448.2[/tex]
[tex]\\ \sf\longmapsto V=\dfrac{448.2}{1.5}[/tex]
[tex]\\ \sf\longmapsto V=298.8mL[/tex]