We must first identify the limiting reactant, and then we calculate the theoretical yield.

We start with the balanced equation.

#color(white)(mmmmmmml)" N"_2 +color(white)(l) "3H"_2 → color(white)(ll)"2NH"_3#

#"MM/g·mol"^"-1": " "28.01color(white)(ll) 2.016color(white)(mml) 17.03#

**(a) Identify the limiting reactant**

We calculate the amount of #"NH"_3# that can form from each reactant.

**Calculate the moles of #"N"_2#**.

#"Moles of N"_2 = 14 color(red)(cancel(color(black)("g N"_2))) × "1 mol N"_2/(28.01 color(red)(cancel(color(black)("g N"_2)))) = "0.500 mol N"_2#

Calculate moles of #"NH"_3# formed from the #"N"_2#

#0.500color(red)(cancel(color(black)("mol N"_2))) × "2 mol NH"_3/(1 color(red)(cancel(color(black)("mol N"_2)))) = "1.00 mol NH"_3#

**Calculate the moles of #"H"_2#**

#"Moles of H"_2 = 12 color(red)(cancel(color(black)("g H"_2))) × "1 mol H"_2/(2.016 color(red)(cancel(color(black)("g H"_2)))) = "5.95 mol H"_2#

Calculate moles of #"NH"_3# formed from the #"H"_2#

#5.95 color(red)(cancel(color(black)("mol H"_2))) × "2 mol NH"_3/(3 color(red)(cancel(color(black)("mol H"_2)))) = "3.97 mol NH"_3#

The limiting reactant is #"N"_2#, because it produces fewer moles of #"NH"_3#.

**(b) Calculate the theoretical yield of #"NH"_3#.**

#"Theoretical yield" = 1.00 color(red)(cancel(color(black)("mol NH"_3))) × "17.01 g NH"_3/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "17.0 g NH"_3#

**(c) Calculate the actual yield of #"NH"_3#**

If % yield = 75 %, then

#"Actual yield" = 17.0 color(red)(cancel(color(black)("g theoretical"))) × "75 g actual"/(100 color(red)(cancel(color(black)("g theoretical")))) = "13 g actual"#