We need to use the stoichiometry of the reaction. The balanced equation is:

#"4NH"_3 + "5O"_2 → "4NO" + "6H"_2"O"#

**Step 1.** Calculate the moles of #"O"_2#.

We can use the Ideal Gas Law:

#color(blue)(|bar(ul(color(white)(a/a) PV = nRT color(white)(a/a)|)))" "#

We can rearrange this to give

#n = (PV)/(RT)#

STP = 1 bar and 0 °C.

∴ # n = (1 color(red)(cancel(color(black)("bar"))) × 12.0 color(red)(cancel(color(black)("L"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K")))) = "0.5284 mol"#

**Step 2.** Calculate the moles of #"NH"_3#.

The balanced equation tells us that 4 mol of #"NH"_3# are required for every 5 mol of #"O"_2#. So,

#"Moles of NH"_3 = 0.5284 cancel("mol O"_2) × ("4 mol NH"_3)/(5 cancel("mol O"_2)) = "0.4227 mol NH"_3#

**Step 3.** Calculate the mass of #"NH"_3#

#"Mass of NH"_3 = 0.4227 color(red)(cancel(color(black)("mol NH"_3))) × ("17.03 g NH"_3)/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "7.20 g NH"_3#