What is the maximum number of grams of ammonia, NH3,?

Molar mass of N₂ = 14.0×2 g/mol = 28.0 g/mol
Molar mass of H₂ = 1.0×2 g/mol = 2.0 g/mol
Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol

Initial number of moles of H₂ = (20.0 g) / (2.0 g/mol) = 10.0 mol
Initial number of moles of N₂ = (160.0 g) / (28.0 g/mol) = 5.71 mol

Balanced equation for the reaction :
N₂ + 3H₂ → 2NH₃
Mole ratio N₂ : H₂ : NH₃ = 1 : 3 : 2

If 10.0 mol H₂ completely reacts, N₂ needed = (10.0 mol) × (1/3) = 3.33 mol < 5.71 mol
N₂ is in excess, and thus H₂ is the limiting reactant/reagent.

Number of moles of H₂ reacted = 10.0 mol
Maximum number of moles of NH₃ obtained = (10.0 mol) × (2/3) = 6.67 mol
Maximum mass of NH₃ obtained = (6.67 mol) × (17.0 g/mol) = 113 g

Ammonia.....

This is a limiting reactant problem. Calculate the amount of product using each reactant. The theoretical yield will be the lesser of the two, and that will tell you the limiting reactant. Use the unit-factor method.

N2(g) + 3H2(g) --> 2NH3(g)
160.g ....20.0g .......?g

160.0g N2 x (1 mol N2 / 28.0g N2) x (2 mol NH3 / 1 mol N2) x (17.0g NH3 / 1 mol NH3) = 194g NH3
20.0g H2 x (1 mol H2 / 2.02g H2) x (2 mol NH3 / 3 mol H2) x (17.0g NH3 / 1 mol NH3) = 112g NH3

112g NH3 is the theoretical yield and H2 is the limiting reactant.

20.0 / 6 = x / 34
x = 113.333 g <<< round as needed