Limiting reactant stoichiometry question? How many grams of ammonia formed during the reaction with 100g of hydrogen and 100g nitrogen?

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 N₂ = (100 g) / (28.0 g/mol) = 3.571 mol
Initial number of moles of H₂ = (100 g) / (2.0 g/mol) = 50 mol

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

If 3.571 mol N₂ completely reacts, H₂ needed = (3.571 mol) × (3/1) = 10.71 mol < 50 mol
H₂ is in excess, and thus N₂ is the limiting reactant/reagent.

No. of moles of N₂ reacted = 3.571 mol
No. of moles of NH₃ formed = (3.571 mol) × (2/1) = 7.142 mol
Mass of NH₃ formed = (7.142 mol) × (17.0 g/mol) = 121 g (to 3 sig. fig.)

The balanced equation for the freaction is:
N2(g) + 3H2(g) → 2NH3(g)
1mol N2 reacts with 3 mol H2 to produce 2 mol NH3
Molar mass N2 = 28g/mol
Mol N2 in 100g = 100g/28g/mol = 3.57 mol N2
This will react with 3*3.57 = 10.71 mol H2
Molar mass H2 = 2g/mol
Mol H2 in 100g = 100g/2g/mol = 50 mol H2
H2 is in excess , thye N2 is limiting

From the equation , 1 mol N2 produces 2 mol NH3
3.57 mol N2 will produce 3.57*2 = 7.14 mol NH3
Molar mass NH3 = 17g/mol
Mass of 7.14 mol NH3 = 17g/mol * 7.14 = 121.4g NH3 produced.

n2+3h2 ->2nh3