What is the theoretical yield in grams?

Molar mass of H₂ = 1.0 × 2 g/mol = 2.0 g/mol
Molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
Molar mass of NH₃ = (14 + 1.0×3) g/mol = 17.0 g/mol

Initial number of moles of H₂ = (1.65 g) / (2.0 g/mol) = 0.825 mol
Initial number of moles of N₂ = (9.56 g) / (28.0 g/mol) = 0.341 mol

3H₂ + N₂ → 2NH₃
Mole ratio H₂ : N₂ : NH₃ = 3 : 1 : 2

If 0.825 mol H₂ completely reacts, N₂ needed = (0.875 mol) × (1/3) = 0.275 mol < 0.341 mol
N₂ is in excess, and thus the limiting reactant/reagent is H₂.

Number of moles of H₂ reacted = 0.825 mol
Maximum number of moles of NH₃ produced = (0.825 mol) × (2/3) = 0.550 mol
Theoretical yield of NH₃ = (0.550 mol) × (17.0 g/mol) = 9.35 g

3H₂+N₂ = 2NH₃
moles of H₂ = mass of H₂/molar mass of H₂ = (1.65 g)/(2.016 g/mol) = 0.818 mol
moles of N₂ = (9.56 g)/(28.01 g/mol) = 0.341 mol
From balanced equation, 3 moles of H₂ reacts with 1 mole of N₂
then 0.818 mol of H₂ will react with (1/3) x 0.818 = 0.273 mol of N₂
However, we have excess N₂, so H₂ is limiting reactant
3 mol of H₂ produces 2 mol of NH₃
then 0.818 mol of H₂ will produce (2/3) x 0.818 = 0.545 mol NH₃
Theoritical yield of NH₃ = mol of NH₃ x molar mass of NH₃ = 0.545 x 17.03 = 9.28 g