The equation for the reaction is
#M_text(r):color(white)(m)80.04color(white)(mmmmmmmmmmmmmm)18.02#
#color(white)(mm)"2NH"_4"NO"_3"(s)" → "2N"_2"(g)" + "O"_2"(g)" + "4H"_2"O(g)"#
A. Volume of nitrogen
#" Moles of N"_2 = 50.0 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × (1 color(red)(cancel(color(black)("mol NH"_4"NO"_3))))/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3))))× ("2 mol N"_2)/(2 color(red)(cancel(color(black)("mol NH"_4"NO"_3)))) = "0.6247 mol N"_2#
You don't give the pressure and temperature at which the nitrogen is formed, so I shall assume STP (1 bar and 0 °C).
The molar volume of an ideal gas at STP is 22.71 L.
#"Volume of N"_2 = 0.6247 "mol N"_2 × "22.71 L N"_2/(1 "mol N"_2) = "14.2 L N"_2#
B. Mass of water
#" Moles of N"_2 = 50.0 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × (1 color(red)(cancel(color(black)("mol NH"_4"NO"_3))))/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3))))× ("4 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol NH"_4"NO"_3)))) = "1.249 mol H"_2"O"#
#"Mass of H"_2"O" = 1.249 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) ="22.5 g H"_2"O" #
