Ca+N-->Ca3N2 -What is limiting reagent? -Mass of calcium nitride formed when 50.0g of calcium reacts with 50.0g of Nitrogen? -Excess?

The chemical formula for nitrogen gas is N₂, but NOT N.

Molar mass of Ca = 40.08 g/mol
Molar mass of N₂ = 14.01 × 2 g/mol = 28.02 g/mol
Molar mass of Ca₃N₂ = (40.08×3 + 14.01×2) g/mol = 148.26 g/mol

Initial number of moles of Ca = (50.0 g) / (40.08 g/mol) = 1.248 mol
Initial number of moles of N₂ = (50.0 g) / (28.02 g/mol) = 1.784 mol

Balanced equation for the reaction :
3Ca + N₂ → Ca₃N₂
Mole ratio Ca : N₂ : Ca₃N₂ = 3 : 1 : 1

When 1.248 mol Ca completely reacts, N₂ needed = (1.248 mol) × (1/3) = 0.416 mol < 1.784 mol
N₂ is in excess, and thus the limiting reagent is Ca.

Number of moles of Ca (completely) reacted = 1.248 mol
Number of moles of N₂ needed = (1.248 mol) × (1/3) = 0.416 mol
Number of moles of Ca₃N₂ produced = (1.248 mol) × (1/3) = 0.416 mol

Mass of N₂ reacted = (0.416 mol) × (28.02 g/mol) = 11.7 g
Mass of N₂ left unreacted = (50.0 - 11.7 g) = 38.3 g

Mass of Ca₃N₂ produced = (0.416 mol) × (148.26 g/mol) = 61.7 g
OR: Mass of Ca₃N₂ produced = Total mass of reactants reacted = (50.0 + 11.7) g = 61.7 g